PHYSICO-CHEMICAL 
CALCULATIONS 


BY 

JOSEPH  KNOX,   D.Sc. 

LECTURER    IN    CHEMISTRY,    UNIVERSITY   OF   ABERDEEN 


NEW  YORK 

D.    VAN    NOSTRAND    COMPANY 

25   PARK  PLAGE 

1916 


Kb 


PREFACE 

THIS   collection    of    physico-chemical   problems    is 
based  on  Abegg    and    Sackur's   "  Physikalisch- 
Chemische  Rechen-aufgaben "  (Sammlung  Goschen). 

The  original  intention  was  simply  to  translate  the 
German  book,  which  consists  of  a  short  summary  of  the 
laws  and  formulae  used  in  the  problems,  and  fifty-two 
typical  problems,  with  full  solutions.  With  the  consent 
of  the  late  Professor  Abegg  and  of  Dr.  Sackur,  however, 
I  decided  to  arrange  the  subject-matter  in  chapters  deal- 
ing with  the  main  subdivisions  of  physical  chemistry, 
and  to  write  a  short  introduction  to  each  chapter,  deal- 
ing with  the  theory  involved  in  the  problems.  Most  of 
the  problems  in  the  "  Kechenaufgaben  "  have  been  re- 
tained, a  good  many  additional  solved  problems  have  been 
introduced,  and  a  collection  of  problems  for  solution  (with 
answers)  has  been  added  at  the  end  of  each  chapter. 
The  size  of  the  book  has  thus  been  more  than  doubled. 

Most  of  the  problems  have  been  taken  direct,  or  with 
slight  modification,  from  the  original  literature. 

I  shall  be  grateful  to  have  any  errors  in  the  text 
pointed  out  to  me. 

I  would  take  this  opportunity  of  expressing  my  in- 
debtedness to  the  late  Professor  Richard  Abegg,  and  to 
Dr.  Otto  Sackur,  of  the  University  of  Breslau,  for  their 
kindness  in  allowing  me  to  make  use  of  their  "  Kechen- 
aufgaben "  as  the  basis  of  this  book. 

J.  K. 

ABERDEEN, 

December,  1911.  , 

362452 


CONTENTS 

PAGE 

CHAPTER  I. 

Gas  Laws—  Gaseous  Dissociation — Osmotic  Pressure — Examples 

— Problems  for  Solution 1 

CHAPTER  II. 

Density  and  Specific  Volume  of  Solids,  Liquids,  Liquid  Mixtures 

and  Solutions— Examples — Problems  for  Solution ...      14 

CHAPTER  III. 

Specific  and   Molecular  Refractivity — Examples — Problems  for 

Solution 19 

CHAPTER  IV. 

Molecular  Weight  from  Lowering  of  Vapour- pressure — Influence 
of  Temperature  on  Vapour-pressure — Examples — Molecular 
Weight  from  Lowering  of  Freezing-point — Molecular  Lower- 
ing of  Freezing-point  from  Latent  Heat  of  Fusion — Molecular 
Weight  from  Elevation  of  Boiling-point — Molecular  Eleva- 
tion of  Boiling-point  from  Latent  Heat  of  Evaporation — 
Examples — Problems  for  Solution  - 24 

tJHAPTER  V. 

Surface  Tension,  Molecular  Weight  and  Degree  of  Association  of 

Liquids— Examples — Problems  for  Solution  -  41 

CHAPTER  VI. 
Thermochemistry — Examples — Problems  for  Solution  45 

CHAPTER  VII. 

Velocity  of  Reaction — Monomolecular  Reaction — Bimolecular  Re- 
action— Examples— Problems  for  Solution     ....      53 
vii 


viii        PHYSICO-CHEMICAL  CALCULATIONS 


CHAPTER  VIII. 

Law  of  Mass  Action  —  Equilibrium-constant  —  Influence  of 
Temperature  on  Equilibrium-constant — Affinity,  Change  of 
Free  Energy  or  Maximum  Work  of  a  Reaction — Partition 
Law — Solubility  of  Gases — Examples — Problems  for  Solution  62 

CHAPTER  IX. 

Ohm's  Law — Heating  Effect  of  Current — Faraday's  Laws — 
Examples — Specific,  Equivalent,  and  Molecular  Conductivity 
of  Electrolytes — Degree  of  Dissociation — Dissociation-con- 
stant —  Examples  —  Transport  Numbers — Examples — Solu- 
bility-product —  Examples  —  Hydrolysis  —  Examples  —  Pro- 
blems for  Solution  ---------  100 


CHAPTER  X. 

Electromotive  Force — Electrode  Potential — Normal  Potential — 
Concentration  Cells  —  Electromotive  Force  of  Galvanic 
Elements — Diffusion  Potential — Oxidation-reduction  Poten- 
tial— Affinity,  or  Maximum  Work  of  a  Reaction  in  a  Galvanic 
Element  —  Gibbs-Helmholtz  Equation  —  Examples  —  Pro- 
blems for  Solution  -  -  -  -  •  -  -  -  149 

CHAPTER  XL 
Diffusion — Examples — Radio-activity — Examples         -        -        -    182 

INDEX 187 


PHYSICO-CHEMICAL 
CALCULATIONS 

CHAPTBE  I 
GAS  LAWS— GASEOUS  DISSOCIATION— OSMOTIC  PKESSUBE 

Gas  Laws 

THE  relation  between  the  pressure,  volume  and  tempera- 
ture  of   any  given  mass  of  gas  is  expressed  by  the 
equation 

/  v  Pf>       P'v' 
•  ~T  = '  "2*" 

where  P  and  v  are  the  pressure  and  volume  corresponding  to 
the  absolute  temperature  T  and  P'  and  v'  the  pressure  and 
volume  corresponding  to  the  absolute  temperature  T.  The 
absolute  temperature  is  equal  to  273  +  t,  where  t  is  the  tem- 
perature centigrade.  The  pressure  and  volume  may  be  ex- 
pressed in  any  units. 

This  relation  may  also  be  expressed  in  the  form 

Pv 

jp  =  constant,  or  Pv  =  constant  x  T. 

The  value  of  the  constant  varies  with  the  units  of  pressure 
and  volume  and  with  the  mass  of  gas  considered.  For  the 
gram-molecular  quantity  of  all  gases,  however,  the  value  of 
the  constant  is  the  same.  For  a  gram-molecule  of  a  gas  the 
equation,  therefore,  becomes 

Pv  =  RT, 
and  for  n  gram-molecules 

(2)  Pv  =  nET  (Boyle,  Gay-Lussac,  Avogadro). 

1 


2  GAS. LAWS— OSMOTIC  PEESSUEE 

P  is  the  pressure  of  the  gas,  n  th'e  number  of  gram -molecules 
or  moles  present  in  the  volume  v,  R  the  gas -constant  and  T 
the  absolute  temperature.  The  numerical  value  of  R  varies 
with  the  units  of  pressure  and  volume  adopted.  If  we  take 
the  atmosphere  as  the  unit  of  pressure  and  the  litre  as  the 
unit  of  volume,  we  obtain  the  corresponding  value  of  R  from 
the  fact  that  one  gram- molecule  of  any  gas  (n  =  1)  occupies 
22-4  litres  at  0°  C.  (273°  absolute)  and  under  a  pressure  of  one 
atmosphere.  Then,  from  (2), 

09-J. 

1  x  22-4  =  1  x  E  x  273  .-.  R  =44r  =  0-08204. 

410 

If  the  pressure  is  given  in  millimetres  of  mercury  and  the 
volume  in  cubic  centimetres,  the  former  must  be  divided  by 
760  to  convert  it  to  atmospheres  and  the  latter  by  1000  to 
convert  it  to  litres  before  this  value  of  R  can  be  used. 

If  the  total  pressure  P  of  a  mixture  of  gases  is  replaced  by 
the  partial  pressure  p  of  any  of  its  components,  the  same 
equations  hold  for  each  individual  gas  in  the  mixture,  of  which 
the  total  volume  is  v  (Dalton's  law). 

Osmotic  Pressure 

For  the  osmotic  pressure  TT  of  a  dilute  solution  of  volume 
v,  equations  exactly  analogous  to  the  gas-equations  hold, 
namely, 

f   ,    irV          ?rV 

(3)  -y    •-     -7jir> 

and 

(4)  TTV  =  nRT. 

TT  is  the  osmotic  pressure,  n  the  number  of  gram-molecules 
of  dissolved  substance  in  the  volume  v  of  solution,  T  the, 
absolute  temperature  and  R  a  constant,  the  value  of  which 
is  the  same  as  in  the  gas-equation  and  numerically  equal  to 
0 '08204  if  TT  is  measured  in  atmospheres  and  v  in  litres* 
The  last  equation  may  be  written  in  the  form  : 

a* 

where  c  is  the  concentration  of  the  dissolved  substance,  (e.g% 
gram-molecules  per  litre,  if  the  above  units  are  adopted). 


GAS  LAWS— EXAMPLES  3 

Gas  Laws — Examples 

PROBLEM  1. — At  £'  =  10°C.  and  under  a  pressure  of  P  ••= 
650  mm.  of  mercury  a  certain  mass  of  hydrogen  occupies  a 
volume  of  v'  =  200  c.c.  What  volume,  v,  will  it  occupy  at 
t  =  0°  C.  and  P  =  760  mm.  pressure  ? 

SOLUTION  1. — Substituting  the  numerical  values  in  equa- 
tion (i)  we  obtain 

760  xv     650  x  200 


273      283 
650  x  200  x  273 
760  x  283 


-  165  c.c. 


PROBLEM  2.  —  What  volume  will  a  =  1  gram  of  oxygen 
occupy  at  a  temperature  t  =  100°  C.  and  under  a  pressure 
P  =  740  mm.  of  mercury  ? 

SOLUTION  2.  —  If  M  is  the  molecular  weight  of  oxygen,  the 

number  of  gram-molecules  present  is  n  =  -^.     Equation  (2), 

therefore,  becomes 

a  aET 

Pv  =       RT,  and  v  = 


740 
P  =  ^atmospheres,  M  =  32,  E   =  Q'08204,  T  =  273  +  t 

=  373.     Substituting  these  values  in  the  above  equation  we 
obtain 

1  x  0-08204  x  373  x  760 
v  =  ~  32  x  740  -  °'9822  htres- 

PROBLEM  3.  —  The  pressure  of  the  atmosphere  at  the  surface 
of  the  earth  is  equal  to  the  weight  of  a  column  of  mercury 
76  cms.  high.  With  increasing  height  the  density  of  the  air 
diminishes  according  to  Boyle's  law.  What  is  the  density  of 
the  air  at  the  heights  \  =  1000  metres,  and  h2  =  3000  metres, 
if  the  density  SQ  at  the  surface  of  the  earth  =  0-00129? 
(t  =  0°  C.). 

SOLUTION  3.  —  According  to  Boyle's  law  the  volume  of  a 
gas  at  constant  temperature  is  inversely  proportional  to  the 
pressure  on  it  ;  its  density,  i.e.  the  mass  contained  in  unit 
volume,  is,  therefore,  directly  proportional  to  the  pressure.  Let 
he  pressure  of  the  air  at  the  height  h  metres  =  P,,  and  its 


4  GAS  LAWS— EXAMPLES 

density  =  s.     Then,  since  s  is  proportional  to  the  pressure 

(1)  s  =  kPt. 
Similarly  at  the  height  h  +  dh, 

(2)  s  +  ds  =  kPs+dt. 

The  difference  of  the  pressures  Pt+df  and  Pg  is  equal  to  the 
weight  of  a  column  of  air  of  height  dh  (the  difference  in  the 
two  heights),  and  of  unit  area  of  cross-section,  since  the  pres- 
sure of  a  mass  of  gas  is  defined  as  equal  to  the  weight  on 
unit  area.  Accordingly  we  have 

From  (1),  (2)  and  (3)  it  follows  that 

ds  =  -  k.s. dh,  or  d log,  s  =  -  k  .  dh, 
and,  after  integration, 

(4)  log,  s  =  -  kh  +  constant. 

At  the  height  h  =  0  we  have  s  =  s0,  the  density  of  the  air 
at  the  surface  of  the  earth,  and  (4)  becomes 

log,  s0  =  constant, 
and,  therefore, 

(5)  loge  s  =  -  kh  4-  loge  s0. 

From  (1 )  it  follows  that  k  is  the  density  of  air  under  unit 
pressure.  The  value  of  k  can  be  calculated  from  the  data 
given  in  the  problem,  i.e.  from  the  value  of  the  density  s0  at 
a  pressure  P  of  76  cms.  of  Hg. 

Using  the  cm.  and  the  gram  as  our  units  P  =  76  x  13-6 
grams  per  sq.  cm.  (13'6  is  the  density  of  Hg),  and  s0  = 
0-00129  gram  per  c.c.,  therefore, 

(V001  2Q 
k  =  7g  x  136  =0-00000125  =1-25  x  1Q-C. 

Then  from  (5)  we  get,  for  the  heights  ht  =  1000  m.  = 
105  cms.,  and  h2  =  3000  m.  =  3  x  105  cms., 

log.  SL  =  -  1-25  x  10~°  x  ftj  +  loge  0-00129, 
=    -  1-25  x  10  -«  x  105  +  log,  0-00129, 

and,  changing  to  common  logarithms, 


GAS  LAWS—  EXAMPLES  5 

log  .<?!  =  -  0'4343  x  0-125  +  log  0  00129, 
=  -  0-0543  +  (0-111  -  3) 
=  0-057  -  3 
.'.  Sj  =  0*00114. 

Iogas2  =  -  1-25  x  10~6  x  hz  +  loge  0-00129, 
=  -  1-25  x  3  x  10  -1  +  log,  0-00129, 
log  $2=  -  0-163  +  (0-111  -  3)  =  0-947  -  4, 
.-.  s2  =  0*000885. 

If  the  densities  at  the  heights  7^  and  7&2  are  referred  to  the 
density  at  the  earth's  surface  as  unity,  then 

0  00114 
si  =  0^00129  =  °'884' 

0-000885 
and  S2  =  0-00129   ==  °'686- 

PROBLEM  4  (of.  preceding  problem).  —  The  volume  v  of  a 
balloon,  which  is  open  below,  is  1400  cubic  metres.  What  is 
its  carrying-capacity  B  at  the  heights  \  =  1000  m.  and  h2  = 
3000  m.  when  filled  (a)  with  hydrogen,  (b)  with  coal  gas,  if 
the  specific  gravity  of  coal  gas  relative  to  air  is  0'43,  and  that 
of  hydrogen  0*0695  ?  (c)  -What  must  be  the  volume  of  a 
balloon  filled  with  hydrogen,  so  that  its  carrying-capacity  may 
be  equal  to  that  of  the  1400  c.m.  balloon  filled  with  coal 
gas? 

SOLUTION  4.  —  The  carrying-capacity  of  a  balloon  can  be 
calculated  from  the  principle  of  Archimedes.  The  balloon 
floats  in  its  position  of  equilibrium  when  the  weight  of 
balloon  +  gas  +  car  with  contents  is  equal  to  the  weight  of 
the  air  displaced  by  them.  For  a  first  approximation  the 
volume  of  the  material  of  the  car  may  be  neglected  in  com- 
parison with  the  volume  of  the  baUoon. 

If  v  is  the  volume  of  the  completely  filled  balloon,  d  the 
density  of  the  balloon  gas  at  the  height  h,  and^s  the  density 
of  the  air  at  the  same  height,  then  the  weight  of  the  balloon 
gas  =  v  .  d,  and  that  of  the  displaced  air  =  v  .  s.  Therefore 
the  carrying-capacity  B  of  the  balloon  =  v  (s  -  d). 

Boyle's  law  holds  both  for  the  balloon  gas  and  for  air; 

accordingly  -^  =S-1>    where  dv    si   are  the   densities  '  at   the 


height  hlt  and  d^  s0  the  densities  at  the  earth's  surface.    The 
carrying-capacity  Bl  at  the  height  7^,  is,  therefore, 


GAS  LAWS— EXAMPLES 


and  at  the  height  h^ 


-A) 

s0/ 


(a)  For  hydrogen  —£.    =  0-0695,    therefore,    since    v  = 

so 

1400  x  106c.c.  and  sl  =  1-14  x  10 -3  (of.  Problem  3),  for  ^ 
=  1000  m.  =  105  cm. 

B!  =  1400  x  10°  x  1-14  x  10  ~3  x  0'93  gm.  =  1484  kg., 
and,  similarly, 

B2  =  1400  x  106  x  0-885  x  10  -  3  x  0'93  gm.  =  1152  kg. 

(b)  For  coal  gas  A  =  0'43,  therefore 

so 
B!  =  1400  x  106  x  1-14  x  10  ~  3  x  0*57  gm.  =  910  kg., 

Bo  =  1400  x  106  x  0-885  x  10  -  3  x  0-57  gm.  =  706  kg. 

(c)  Let  the  volume  of  the  hydrogen  balloon  be  vm  then  its 
carrying-capacity  B  =  vxs(l  -  —  Y    Similarly,  for  a  coal  gas 

balloon  of  the  same  carrying-capacit    we  have 


dn,  dc  and  s  are  the  densities  at  any  given  height  of  hydrogen, 
coal  gas  and  air  respectively.  From  these  two  equations  we 
obtain 


~S 

and,  therefore, 


PROBLEM  5  (cf.  preceding  problem).  —  If  the  balloon  of 
1400  c.m.  capacity  is  filled  completely  at  the  earth's  surface 
(a)  with  hydrogen,  (b)  with  coal  gas,  how  many  grams  of 
gas  are  in  each  case  lost  when  it  ascends  to  the  heights  hl  = 


GASEOUS  DISSOCIATION— EXAMPLES  7 

1000  ro.  and  h>2  =  3000  m.  ?  How  many  kilograms  of  ballast 
must  be  thrown  out  so  that  the  completely  filled  balloon 
may  rise  from  its  equilibrium  position  at  the  height  /^  = 
1000  m.  to  that  at  \  =  3000  m.,  (c)  when  filled  with  hydro- 
gen, (d)  when  filled  with  coal  gas  ? 

SOLUTION  5. — The  weight  of  gas  lost  on  ascending  to  the 
height  /ix  is  OJ  =  v  (dQ  -  d^  grams,  and  on  ascending  to  the 
height  h«  it  is  G2  =  v  (dQ  -  d2)  grams,  where  dQ9  dl  and  d»  are 
the  densities  of  the  gas  at  the  earth's  surface,  at  the  height  /^ 
and  at  the  height  h2  respectively. 

(a)  For  hydrogen 

d0  -  dl  =  0-0695  (0-00129  -  O'OOIH) 

=  0-0695  x  0-00015  -»  1-04  x  10  ~5, 
and  d0  -  d<2  =  0-0695  (0-00129  -  0-000885) 

-  0-0695  x  0-000405  ='2-81  x  10  ~5, 
therefore  the  loss  of  weight 

Gx  =  1400  x  106  x  1-04  x  10  -  6  gm.  =  14-56  kg., 
and  G2  =  1400  x  106  x  2-81  x  10  ~  5  gm.  =  39-3  kg. 

(b)  Similarly,  for  coal  gas 

Gx  =  1400  x  106  x  6-45  x  10  ~5  gm.  =  90-3  kg., 
and  G2  =  1400  x  106  x  T74  x  10  - 5  gm.  =  243-5  kg- 
The  weight  of  ballast  which  must  be  thrown  out  is  equal 
to   the   difference  between    the   carrying-capacities   at    the 
heights  hj_  and  /&2,   and  is,  therefore, 

(c)  for  hydrogen,  1484  -  1152  =  332  kg. 

(d)  for  coal  gas,      910  -     706  =  204  kg. 

The  throwing  out  of  a  given  weight  of  ballast  has,  there- 
fore, more  effect  on  the  coal-gas  balloon  than  on  the  hydrogen 
balloon. 

Gaseous  Dissociation — Examples 

PROBLEM  6. — At  90°  C.  the  vapour-density  of  nitrogen  per- 
oxide (N204),  is  24-8  (referred  to  H  =  1).  Calculate  the 
degree  of  dissociation  into  N02  molecules  at  this  temperature. 

SOLUTION  6. — Let  D  be  the  theoretical  vapour-density  if  no 
dissociation  occurred,  and  d  the  observed  vapour-density.  If 
one  molecule  of  the  dissociating  substance  gives  on  dissociating 
n  simpler  molecules,  and  if  a  is  the  degree  of  dissociation,  then 
at  equilibrium  1  -  a  undissociated  molecules  and  no.  simpler 
molecules  are  present  for  every  molecule  of  undissociated 
substance  initially  present.  The  total  number  of  molecules 
is,  therefore,  1  -  a  +  na  =  1  +  a(n  -  1).  The  volume 


8     •      GASEOUS  DISSOCIATION—  EXAMPLES 

occupied  by  the  gram-molecular  quantity  of  undissociated 
substance  is,  therefore,  1  +  a(n  -  1)  times  as  great  as  it 
would  have  been,  had  no  dissociation  occurred.  But  the 
density  of  a  given  mass  is  inversely  proportional  to  the 
volume,  therefore, 

D     i  +  a(n  _  i) 

-j=     ~T- 

and 

_  (D  -  d) 
=  d(n  -  1)  * 
In  the  dissociation  of  N.2O4  according  to  the  equation 

N204  =  2N02, 
n  =  2,  D  =  N2°4  =  ^  =  46,  therefore 

2>  z> 

(D  -d)      46  -  24-8 


PROBLEM  7.  —  When  a  =  5  grams  of  ammonium  carbarn  ate 
NH4CO2NH2,  is  completely  vaporised  at  t  =  200°  C.,  it  occu- 
pies a  volume  v  =  7'66  litres  under  a  pressure  P  of  740  mm. 
of  mercury.  Calculate  the  degree  of  dissociation  according 
to  the  equation 

NH4C02NH2  =2NH3  +  CO,. 

SOLUTION  7.  —  If  Mis  the  molecular  weight  of  ammonium 
carbamate,  a  grams  =  —  gram-molecules.  Let  a  be  the  de- 

gree of  dissociation.  Then  since  one  molecule  on  dis- 
sociating gives  2a  molecules  of  NH3  and  a  molecules  of  CO2, 
whilst  1  -  a  molecules  of  undissociated  substance  remain, 
the  total  number  of  molecules  at  equilibrium  derived  from  1 
molecule  of  undissociated  substance  is  1  -  a  +  3a  =  1  +  2a 

molecules.     From  —  molecules  of  undissociated  substance 
M 

there  are,   therefore,    derived  ~  (1  +  2a)  molecules.      The 

equation 

Pv  =  nRT, 
therefore,  becomes 

Pv  =        l  + 


OSMOTIC  PRESSURE—  EXAMPLE  9 

M  =  78  and  P  =    -  —   atmos.     Substituting    the    numerical 

i  t>U 
values  we,  therefore,  obtain 


x  ?.66  =   5  (1  +  2a)  x  0.08204  x  473, 
7bO  7o 

.'.a  =  0*999. 

The  compound  is,  therefore,  completely  dissociated. 

Osmotic  Pressure  —  Example 

PROBLEM  8.  —  A  solution  of  glucose  containing  a  =  36 
grams  per  litre  gave  an  osmotic  pressure  TT  =  4'  77  atmo- 
spheres at  t  =  25°  C.  What  is  the  molecular  weight  M  of 
glucose  ? 

SOLUTION  8.  —  If  a  is  the  weight  of  substance  in  solution, 
and  M  its  molecular  weight,  the  number  of  dissolved  mole- 

cules is  n  =   T,.    Equation  (4)  therefore  becomes 


and 

aET 


M 


irV 


Substituting  the   numerical  values  in  this    equation   we 
obtain 

36  x  0-08204  x  (273  +  25) 

4-77  x  1  =  I84'5- 

Problems  for  Solution 

Gas  Laws 

PROBLEM  9. — At  0°  C.  and  760  mm.  pressure  the  volume 
occupied  by  1  gram-molecule  of  oxygen  is  22-4  litres.     Cal- 
culate the  numerical  value  of  the  gas-constant  E  when  the 
units  of  energy  are  (1)  the  litre-atmosphere,  (2)  the  gram- 
centimetre,  (3)  the  erg,  (4)  the  gram-calorie,  (5)  the  joule. 
Given :     density    of    mercury  =  13*59,     gravity-constant  = 
981,  1  gram-calorie  =  42720  gram-centimetres  =  4-183  joule. 
Ans.  (1)  0-08204,  (2)  84720,  (3)  0-8312  x  108, 
(4)  1-984,  (5)J^301.      ' 

PROBLEM  10. — At  10°  C.   and  under  a  pressure  of  1000 


10  GAS  LAWS— PROBLEMS 

grams  per  square  centimetre,  a  certain  mass  of  carbon 
dioxide  occupies  a  volume  of  30  cubic  inches.  What  volume 
will  it  occupy  at  20°  C.  and  under  a  pressure  of  850  grams 
per  square  centimetre? 

Ans.  36*55  cubic  inches. 

PROBLEM  11. — An  open  vessel  at  a  temperature  of  10°  C.  is 
heated  at  constant  pressure  to  400°  C.     What  fraction  of  the 
weight  of  air  originally  contained  in  the  vessel  is  expelled  ? 
Ans.  0-5794. 

PROBLEM  12. — If  the  vessel  in  the  preceding  problem  had 
been  closed,  and  originally  contained  air  at  atmospheric  pres- 
sure, what  pressure  would  have  been  developed  on  heating  to 
400°  C.? 

Ans.  2-378  atmos. 

PROBLEM  13. — From  a  porcelain  bulb  of  volume  197'8  c.c. 
filled  with  air,  169'1  c.c.  of  air,  measured  at  10°  C.,  were 
expelled  on  heating  from  12°  C.  to  t°  C.  The  pressure 
throughout  the  experiment  was  747  mm.  Neglecting  the 
expansion  of  the  bulb,  calculate  t. 

Ans.  t  •-=  1772°  G. 

PROBLEM  14. — A  bulb  of  111*5  c.c.  capacity  connected  with 
a  mercury  manometer  was  heated  to  100°  C.  whilst  open  to 
the  atmosphere.  When  the  temperature  had  become  con- 
stant, the  atmospheric  connexion  was  closed,  and  a  capsule 
containing  0'0448  gram  of  acetone  dropped  into  the  heated 
bulb.  The  volume  was  kept  constant  by  increasing  the  pres- 
sure by  raising  the  reservoir  of  the  manometer.  The  final 
difference  of  level  was  16' 1  cms.  Calculate  the  molecular 
weight  of  acetone. 

Ans.  57-65. 

PROBLEM  15. — Half  a  gram  of  an  organic  compound  of 
empirical  formula  CH2O  gave  327 '6  c.c.  of  vapour  at  200°  C., 
and  750  mm.  pressure.  What  is  the  molecular  formula  of 
the  compound  ? 

Ans.  C2H4O2. 

PROBLEM  16. — At  "17°  C.  40  grams  of  electrolytic  gas  are 
contained  in  a  60  litre  gas  holder.  What  are  the  partial 
pressures  of  hydrogen  and  oxygen  ? 

Ans.  ^  =  0-8812  atmos.,  p0t)  =  0-4406  atmos. 

PROBLEM  17. — The  composition  of  air  by  weight  is  23  per 


GASEOUS  DISSOCIATION— PEOBLEMS         11 

cent,  oxygen  and  77  per  cent,  nitrogen.  What  are  the  partial 
pressures  of  oxygen  and  nitrogen  in  a  vessel  of  1  litre  capacity 
•which  eontains  2  grams  of  air  at  15°  0.  ? 

Ans.  pQ^  =  0-3397  atmos.,  p^  =  1*299  atmos. 

PROBLEM  18.- — What  weights  of  hydrogen,  oxygen,  and  nitro- 
gen are  contained  in  10  litres,  measured  at  18°  C.  and  750  mm., 
of  a  gaseous  mixture,  the  volumetric  composition  of  which  is 
H2  =  10  per  cent.,  O2  =  15  per  cent.,  N2  =  75  per  cent.? 
Ans.  H2  =  0-0826,  O2  =  1-985,  N2  =  8 -684  grams. 

Gaseous  Dissociation 

PROBLEM  19. — At  26°  C.  the  vapour-density  of  nitrogen 
peroxide  is  38,  referred  to  hydrogen  as  unit.  Calculate  the 
proportion  of  N2O4  molecules  to  NO2  molecules  in  the  vapour 
at  this  temperature. 

Ans.  N2O4/N02  =  1-869. 

PROBLEM  20. — The  vapour-density  of  bromine  (at.  wt.  =  80) 
at  1000°  C.  is  76-94  (H  =  1).    What  is  the  degree  of  dissociation 
of  diatomic  into  monatomic  molecules  at  this  temperature  ? 
Ans.  4  per  cent. 

PROBLEM  21. — At  70°  C.  and  under  atmospheric  pressure 
N«,04  is  65-6  per  cent,  dissociated  into  N02.  What  volume 
will  10  grams  of  N204  occupy  under  these  conditions? 

Ans.  5-065  litres. 

PROBLEM    22. — The    vapour-density   of    nickel   carbonyl, 
Ni  (C0)4,  is  83-3  at  63°  C.  and  70-8  at  100°  C.    Calculate  the 
percentage  dissociation  at  these  two  temperatures  (Ni  =  58). 
Ans.  0*7  per  cent,  and  6'66  per  cent. 

PROBLEM  23. — At  200°  C.  and  atmospheric  pressure  the 
vapour-density  of  PC15  is  70  (H  =  1)  Calculate  the  degree 
of  dissociation  of  the  PC15  vapour.  What  are  the  partial 
pressures  and  concentrations  in  gram-molecules  per  litre  of 
PC15,  PC13  and  C12,  when  a  gram-molecule  of  PC15  is  heated 
under  atmospheric  pressure  to  200°  C.  ?  (P  =  31,  Cl  =  35*5). 
Ans.  Dissociation  =  48*9  per  cent. 

Partial  pressures.  PC15  =  0-3431  atmos.,  PC13  and  C12  = 
0-3283  atmos. 

Concentrations.    PC15  =  0-008841,  PC13  and  C12  =  0-008461. 

PROBLEM  24. — When  ammonium  carbamate,  NH^CO^NEL, 


12     .        OSMOTIC  PEESSUEE— PEOBLEMS 

is  heated,  it  dissociates  completely  into  2NH3  +  CO2.  What 
volume  will  be  occupied  by  the  gaseous  products  from  7'8 
grams  of  ammonium  carbamate  at  100°  C.  and  740  mm. 
pressure  ? 

Ans.  9-428  litres. 

Osmotic  Pressure 

PEOBLEM  25. — At  10°  C.  the  osmotic  pressure  of  a  solution  of 
urea  is  500  mm.  of  mercury.  If  the  solution  is  diluted  to 
10  times  its  original  volume,  what  is  the  osmotic  pressure  at 
15°  C.  of  the  diluted  solution? 

Ans.  50-89  mm. 

PROBLEM  26. — At  20°  C.  the  osmotic  pressure  of  a  cane  sugar 
solution  is  800  mm.  of  mercury.  What  will  it  be  at  0°  C.  ? 

Ans.  745-4  mm. 

PROBLEM  27. — At  21-8'J  C.  the  osmotic  pressure  of  a  cane 
sugar  solution  containing  68-4  grams  per  litre  was  4-81  atmos. 
Calculate  the  numerical  value  of  the  constant  E  when  the 
units  of  pressure  and  volume  are  the  atmosphere  and  the  litre. 

Ans.  E  =  0-0816. 

PROBLEM  28. — The  osmotic  pressure  of  a  solution  of  0'184 
gram  of  urea  in  100  c.e.  water  was  56  cms.  of  mercury  at 
30°  C.  Calculate  the  molecular  weight  of  urea. 

Ans.  62-03. 

PROBLEM  29. — What  is  the  osmotic  pressure  in  atmospheres 
at  24-2°  C.  of  a  solution  of  glucose  containing  0-5  gram-mole- 
cule per  litre  ? 

.    Ans.  12-2. 

PROBLEM  30.  At  24°  C.  the  osmotic  pressure  of  a  cane 
sugar  solution  is  2-51  atmos.  What  is  the  concentration  of 
the  solution  in  gram-molecules  per  litre  ? 

Ans.  0-103. 

PROBLEM  31. — At  25*1°  C.  the  osmotic  pressure  of  a  solution 
of  glucose  containing  18  grams  per  litre  was  2-43  atmos. 
Calculate  the  numerical  value  of  the  constant  E  when  the 
unit  of  energy  is  the  gram-centimetre. 

Ans.  E  =  84300.  1*^-° 

PROBLEM  32. — At  18°  C.  a  0-5  N  -  NaCl  solution  is  74-3 
per  cent,  electrolytically.  dissociated.      What  would  be  the 
osmotic  pressure  of  the  solution  in  atmosphereiHit  18°  C.  ? 
Ans.  20-79. 


OSMOTIC  PEESSUEE— PEOBLEMS         .     13 

PROBLEM  33. — A  solution  containing  3  gram-molecules  of 
cane  sugar  per  litre  was  found  by  the  plasmolytic  method  to 
be  isosmotic  with  a  solution  of  potassium  nitrate  containing 
1*8  gram -molecules  per  litre.  What  is  the  degree  of  dis- 
sociation of  the  potassium  nitrate? 
Ans.  0-67. 

PROBLEM  34. — A  solution  containing  1/9  gram-molecules  of 
calcium   chloride   per  litre  is   isosmotic   with  a  solution  of 
glucose  containing  4/05  gram-molecules  per  litre.     What  is 
the  degree  of  dissociation  of  the  calcium  chloride  ? 
Ans  0-566. 

PROBLEM  35. — In  a  solution  containing  1  gram-molecule 
of  potassium  bromide  in  8  litres,  the  salt  is  82  per  cent,  dis- 
sociated at  25°  C.  What  is  the  osmotic  pressure  of  the 
solution  at  this  temperature  ? 

Ans.  5 '56  atmos. 


CHAPTBE  II 

DENSITY  AND  SPECIFIC  VOLUME  OF  SOLIDS,  LIQUIDS, 
LIQUID  MIXTUBES  AND  SOLUTIONS 

Definitions 

THE  density  d  of  a  substance  at  a  given  temperature  t  is 
the  weight  of  a  given  volume  of  it  compared  with  the 
weight  of  the  same  volume  of  water  at  4°  C. ;  or  it  is  the 
mass  of  unit  of  volume  of  the  substance  (grams  per  c.c.). 
This  is  usually  written  d\.  .  I 

The  reciprocal  of  the  density-  the  specific  volume  or  volume 
of  unit  mass,  is  V  =  1/d,  and  in  the  usual  units  gives  the 
volume  in  c.c.  occupied  by  one  gram. 

For  many  liquid  mixtures  and  solutions  the  volume  of  the 
mixture  is  equal  to,  or  approximately  equal  to,  the  sum  of  the 
volumes  of  the  components,  so  that  the  specific  volume,  and, 
therefore,  the  density  of  the  mixture  may  be  calculated  from 
those  of  its  components  by  the  mixture-formula.  Conversely, 
the  approximate  composition  of  the  mixture  may  be  calcu- 
lated from  its  known  specific  volume  or  density  and  those  of 
its  components.  Thus  it  the  specific  volume  of  the  substance 
A  is  FA  and  that  of  the  substance  B  is  FB,  the  specific  volume 
FM  of  a  mixture  of  A  and  B  containing  p  per  cent,  by  weight 
of  A  is  given  by  the  formula 

(i)  100  FM  =  pVA  +  (100  -  p)  FB. 

The  more  closely  chemically  related  the  components  are, 
the  more  accurately,  as  a  rule,  does  the  formula  reproduce 
the  actual  results. 

Density  and  Specific  Volume — Examples 

PROBLEM  36. — The  density  of  a  c  =  0'528  molecular  N-urea 
solution  is  d  =  1*0104.  How  many  grams  of  water  per  gram 
of  urea  does  the  solution  contain  ?  What  is  the  specific 
volume  F  of  urea  in  the  solution  ? 

SOLUTION  36. — A  c  molecular  N-solution  contains  cM  gms. 
per  litre,  if  M  is  the  molecular  weight  of  the  dissolved  sub. 

14 


DENSITY,  SPECIFIC  VOLUME—  EXAMPLES     15 

stance  ;  a  litre  of  the  solution  weighs  1000  d  gms.,  and  there- 
fore contains 

1000  d  -  cM  gms.  of  water. 

The  c  N-urea  solution,  therefore,  contains  per  gm.  of  urea 

1000  d  -  cM 

cM 


The  specific  volume  V  of  urea  in  the  solution  is  the  volume 
in  c.c.  occupied  by  1  gm.  Now  1000  c.c.  of  the  solution 
contain  (1000  d  —  cM)  gms.  of  water  which  occupies  (1000  d 
—  cM)  c.c.,  if  it  is  assumed  that  the  specific  volume  of  the 
water  is  not  changed  by  solution.  The  volume  occupied  by 
cM  gms.  of  urea  is,  therefore,  1000  -  (1000  d  -  oM)  c.c. 
Hence  the  specific  volume  is 

1000  -  1000  d  +  cM 
cM 

1000  -  10104  +  31-7 

31.7  -  =  0-67  c.c.  per  gm. 

PROBLEM  37.  —  What  is  the  normality  of  an  a  =  4  per  cent. 
(by  weight)  cane  sugar  solution,  if  the  specific  volume  of  solid 
cane  sugar  is  V  =  0'615,  and  if  no  change  of  volume  accom- 
panies solution? 

SOLUTION  37.  —  Let  the  density  of  an  a  per  cent,  cane  sugar 
solution  =  d.  Then  a  litre  of  the  solution  weighs  1000  d 


gms.     In  these  1000  d  gms.  there  are  a  x  -     -  10 

JLUU 

ad  gms.  of  cane  sugar.     If  M  is  the  molecular  weight  of  cane 

sugar  the'normality  of  the  solution  is  c  —  -  £-^-!  • 

M. 

d  is  calculated  as  follows  :  the  volume  of  1  litre  of  solution 
is  equal  to  the  sum  of  the  volumes  of  the  dissolved  sugar  and 
of  the  water  used  for  its  solution.  A  litre  of  solution  contains 
10  x  ad  gms.  of  sugar,  of  which  the  volume  is  10  x  a^Fc.c., 
and  (1000  d  -  10  ad)  gms.  of  water,  of  which  the  volume  is 
(1000  d  -  10  ad}  c.c.  Therefore 

1000  -  lOadV  +  1000  d  -  Wad 
and 

1000 
"  10  o(7-  1)4-  1000' 


16    DENSITY,  SPECIFIC  VOLUME— EXAMPLES 

The  normality  of  the  solution  is,  therefore, 

1000  g 
{a(V  -  1)  +  1W\M' 

1000  x  4 

=  (100  -  4  x  0-385)342  =  °'» 9  gram-molecule  per  litre. 

PROBLEM  38. — What  is  the  specific  volume  of  potassium 
hydroxide  in  a  solution  which  contains  1  gram-molecule  of 
KOH  in  1000  grams  of  water,  and  of  which  the  density  is 
d  =  1-052,  if  the  density  of  water  is  taken  =  1-000,  and  if  it 
is  assumed  that  the  specific  volume  of  water  is  not  changed 
by  the  process  of  solution  ? 

SOLUTION  38. — If  M  is  the  molecular  weight  of  KOH, 
then  1000  +  M  grams  of  the  solution  occupy  a  volume  of 

^ —    -  c.c.    The  volume  of  the  water  used  in  making  the 

solution  is  1000  c.c.,  therefore  the  volume  of  the  dissolved 
KOH  is  1QOQ  +  M  _  1000  c.c.  The  specific  volume  of  the 

dissolved  KOH,  i.e.  the  volume  in  c.c.  occupied  by  1  gram 
of  KOH,  is,  therefore, 

1000  +  M  -  1000  d  =  1000  +  56-15  -  1052  = 

Md  56-15  x  1'052 

gram.  The  smallness  of  this  value  makes  it  probable  that 
the  assumption,  that  the  water  does  not  change  its  volume  in 
the  process  of  solution,  is  inadmissible. 

PROBLEM  39. — <^°  for  carbon  disulphide  is  1  -264,  and  for 
ethyl  alcohol  0'7963.  What  is  its  value  for  a  mixture  of 
carbon  disulphide  and  alcohol  containing  79-82  per  cent,  of 
the  former? 

SOLUTION  39. — Since  V  =  l/d  we  may  write  formula  (i)  in 
the  form 

100       p        (100  -  p) 


p  =  79-82,  dA  =  1-264,  dB  =  0-7963,  therefore 
100      79-82        20-18 
dM  ~  1-264  +  0-7963' 

and  dM  =  1-130.     (Observed,  1-122.) 


DENSITY,  SPECIFIC  VOLUME— PROBLEMS    17 

Problems  for  Solution 

PROBLEM  40. — The  density  of  a  2-31  per  cent,  solution  of 
ammonia  is  0*990.  What  is  the  concentration  of  the  solution 
in  gram- molecules  per  litre  ? 

Ans.  1*345. 

PROBLEM  41.— d^  for  a  10  per  cent.  NH4C1  solution  =  1-029, 
for  solid  NH4C1  =  1-536,  and  for  water  =  0-9974.  Calculate 
the  change  of  volume  per  100  grams  solution  in  making  a  10 
per  cent  solution  of  NH4C1.  What  is  the  specific  volume  of 
NH4C1  in  the  solution,  if  that  of  the  water  remains  constant  ? 

Ans.  Expansion  =  0*48  c.c.  per  100  grams  solution.  Sp, 
vol.  NH4C1  =  0-699  c.c.  per  gram. 

PROBLEM  42. — The  density  of  a  4-526  per  cent,  solution  of 
HgCl2  in  water  at  16°  C.  =  1-038,  and  that  of  an  11-88  per 
cent,  solution  in  alcohol  at  16°  C.  =0-8857.  The  density  of 
water  at  16°  C.  =  0;9979,  and  of  alcohol  =  0*7939.  Calcu- 
late the  concentration  of  each  solution,  and  the  specific 
volume  of  HgCl2  in  each.  Assume  no  change  in  the  specific 
volumes  of  the  water  and  alcohol  on  solution. 

Ans.  Cone,  aqueous  solution  =  0-1733  gram-molecule  per 
litre.  Cone,  alcoholic  solution  =  0-3883  gram-molecule  per 
litre. 

Sp.  vol.  in  aqueous  solution  =  0-1575  c.c.  per  gram,  in 
alcoholic  solution  =  0-1635  c.c.  per  gram. 

PROBLEM  43. — For  a  5-18  per  cent,  solution  of  phenol  in 
water  d™  =  1-0042 ;  for  water  d™  =  0-9991.  What  is  the 
concentration  of  the  solution  in  moles  per  litre,  and  in  moles 
phenol  per  mole  water?  What  is  the  specific  volume  of 
phenol  in  the  solution,  if  the  specific  volume  of  the  water  is 
assumed  to  remain  constant  ? 

Ans.  Cone.  =  0-5534  mole  per  litre  =  0'0104&mole  phenol 
per  mole  water. 

Sp.  vol.  =  0-9363  c.c.  per  gram. 

PROBLEM  44.—  d™  for  ethyl  alcohol  =  0'7936,  for  water  = 
3-9991,  and  for  a  50  per  cent,  solution  of  alcohol  in  water  = 
D-9180.  What  is  the  contraction  on  mixing  50  grams  of 
ilcohol  with  50  grams  of  water  at  15°  C.  ?  Assuming  the 
specific  volume  of  the  water  to  remain  constant,  compare  the 
2 


18     DENSITY,  SPECIFIC  VOLUME— PROBLEMS 

specific  volume  of  pure  alcohol  with  that  of  the  alcohol  in  the 
solution. 

Ans.  Contraction  =  4*16  c.c.  Sp.  vol.  of  pure  alcohol  = 
1-260,  of  alcohol  in  the  solution  =  T177  c.c.  per  grain. 

PROBLEM  45.— d1^  for  ethylene  dibromide  =  2-183,  for 
propyl  alcohol  =  0*8066,  and  for  a  solution  of  p  per  cent, 
ethylene  dibromide  in  propyl  alcohol  =  0-8608.  Calculate 
the  value  of  p. 

Ans.  10-10.     (Observed,  10-01.) 

PROBLEM  46  (cf.  preceding  problem). — Calculate  the  density 
at  18-7°  C.  of  a  20-95  per  cent,  solution  of  ethylene  dibromide 
in  propyl  alcohol. 

Ans.  0-9292.     (Observed,  0-9291.) 

PROBLEM  47. — d™  for  carbon  disulphide  =  1-264,  for  ethyl 
alcohol  =  0*7963  and  for  a  50-77  per  cent,  solution  of  carbon 
disulphide  in  alcohol  =  x.     Calculate  the  value  of  x. 
Ans.  0-9804.     (Observed,  0-9718.) 

PROBLEM  48. — d*6'3  for  aniline  =  1*025,  for  ethyl  alcohol 
=  0-8081  and  for  a  p  per  cent,  solution  of  aniline  in  alcohol 
=  0-9763.  Calculate  p. 

Ans.  81-34.     (Observed,  79-24.) 

PROBLEM  49. — d™  for  benzene  =  0-8814,  for  ethyl  alcohol 
=  0-7930.  What  is  its  value  for  a  21-12  per  cent,  solution  of 
benzene  in  alcohol  ? 

Ans.  0-8104.     (Observed,  0-8106.) 

PROBLEM  50  (cf.  preceding  problem). — The  density  of  a 
p  per  cent,  solution  of  benzene  in  alcohol  at  20°  C.  is  0-8604. 
What  is  the  value  of  p  ? 

Ans.  77-94.     (Observed,  79-10.) 


CHAPTEE  III 
SPECIFIC  AND  MOLECULAE  KEFKACTIVITlT 

Definitions 

'"PHE  refractive  index  n  of  a  substance  for  light  of  a  given 
J-      wave-length,  varies  with  the  temperature,  but  the  ex- 
pressions n-^-     (Gladstone    and     Dale),    and    n<)  "     .  — . 
a  n*  +  2     a 

(Lorentz  and  Lorenz),  where  d  is  the  density  of  the  substance 
at  the  temperature  at  which  n  is  measured,  are,  for  a  given 
substance,  almost  constant  and  independent  of  the  tempera- 
ture. This  is  especially  the  case  with  the  latter  expression, 
which  remains  approximately  constant  even  for  different 
states  of  aggregation  of  the  substance  (e.g.  liquid  and  vapour). 

The  value  of  ?Ll — .  or  of  net  ~  .  .  -=  is   called   the   specific 
d  n2  +  2     d 

refractive  power  or  specific  refractivity  of  the  substance.  If 
the  specific  refractivity  is  multiplied  by  the  molecular  weight 

M  of  the  substance,  the  molecular  refractivity  (i)  ^ — ~     ' — 

d 

or  (2)    nn  ~  -  •  -r   is  obtained.     The  molecular  refractivity 
v  }    ri*  +  2     d 

of  a  compound  may  be  calculated  from  the  atomic  refrac- 
tivities  of  its  component  elements.  For  monovalent  elements 
(e.g.  H2,  C12,  Br2)  the  atomic  refractivity  is  practically  con- 
stant and  independent  of  the  nature  of  the  other  elements 
with  which  they  are  united.  For  polyvalent  elements  (e.g. 
O2,  C,  N2)  the  value  of  the  atomic  refractivity  varies  with  the 
mode  of  union  to  the  other  elements  of  the  compound,  but- 
is  constant  for  a  given  mode  of  union.  Thus  in  the  case  of 
sarbon  the  atomic  refractivity  of  a  singly  bound  (saturated) 
carbon  atom  is  different  from  that  of  a  doubly  or  trebly  linked 
carbon  atom.  This  can  be  allowed  for  by  assigning  a  definite 
'  atomic  "  refractivity  to  an  ethylene  (double)  and  to  an 

19 


20  REFKACTIVITY— EXAMPLES 

acetylene  (triple)  bond.  The  molecular  refractivity,  as  cal- 
culated from  the  refractive  index,  may,  therefore,  sometimes 
serve  as  a  guide  to  the  constitution  of  a  compound.  The 
following  are  the  atomic  refractivities  for  the  D  line  (sodium 
light),  calculated  according  to  Lorentz  and  Lorenz's  formula, 
of  the  elements  required  in  the  following  problems.  Carbon 
(singly  bound)  =  2-501,  hydrogen  =  1-051,  oxygen  in  a 
hydroxyl  (-  OH)  group  =  1-521,  oxygen  in  ethers  =  1'683, 
oxygen  in  a  carbonyl  (>C  =  O)  group  =  2-287,  chlorine  = 
5-998,  ethylene  bond  =  1-707,  acetylene  bond  =  2-10. 

Refractivity  of  Mixtures 

The  specific  refractivity  of  a  homogeneous  mixture  or 
solution  may  be  calculated  (as  a  rule  with  close  approxima- 
tion to  the  observed  value)  from  the  specific  refractivities  of 
its  components  by  the  mixture-formula  (cf.  specific  volume). 
Thus  if  nA  is  the  refractive  index  of  substance  A,  nB  that  of 
substance  B  and  n^  that  of  a  mixture  containing  p  per  cent, 
by  weight  of  A,  all  for  light  of  the  same  wave-length,  and  dA, 
ds  and  dm  the  corresponding  densities,  we  have,  using  Glad- 
stone and  Dale's  formula, 

nm  -I  _  nA  -  I     ^_      nB-  1     (100  -  p) 
(3)      du  dA     '  100          ds  100 

or,  using  Lorentz  and  Lorenz's  formula, 

n\-l     _!    =  n*_-_l        P          n\-l     (100  -  p) 
(4)  n*    +  2  *  d        n\  +  2'  100  dA      n\  +  2  '      100  dB    ' 


Conversely,  if  the  specific  refractivites  and  densities  of  the 
mixture  and  its  components  are  known,  the  composition  of 
the  mixture  may  be  calculated. 

Refractivity — Examples 

PROBLEM  51. — For  propionic  acid  d°4  =  1-0158  and  nD,  the 
refractive  index  for  sodium  light,  =  1-3953.  Calculate  the 
molecular  refractivity  of  propionic  acid  by  Lorentz  and 
Lorenz's  formula  and  compare  it  with  the  value  calculated 
from  the  atomic  refractivities  given  above. 

SOLUTION  51. — M,  the  molecular  weight  of  propionic  acid, 
=  74.  Substituting  the  numerical  values  in  formula  (2)  we 
obtain 

(1'3953)2  -  1         74 
(l-3953)2-f  2      1-0158        74 
for  the  molecular  refractivity. 


REFRACTIVITY— EXAMPLES,  PEOBLEMS       21 
From  the  atomic  refractivities  we  obtain  for 

CH3  •  CH2  •  G{ 

\OH 

3  carbon  atoms  =  3  x  2'501  =  7*503 

6  hydrogen  atoms  =  6  x  1-051  =  6-306 

1  hydroxylic  oxygen  atom  =  1  x  1-521  =  1-521 

1  carbonyl  oxygen  atom  =  1  x  2-287  =  2 -287 

Molecular  refractivity  =  17-62. 

PROBLEM  52. — d™  for  ether  =  0-7208,  for  ethyl  alcohol  = 
0'7935  and  for  a  mixture  of  ether  and  alcohol  containing^ 
per  cent,  of  alcohol  =  0'7389<  At  20°  C.  the  refractive  indices 
for  sodium  light  are,  for  ether  1-3536,  for  alcohol  1-3619,  and 
for  the  mixture  1-3572.  Calculate  the  value  of  p,  using  the 
Gladstone  and  Dale  formula. 

SOLUTION  52. — Substituting  the  numerical  values  in  equa- 
tion (3)  we  obtain 

(1-3572  -1)  _  (1-3619  -  1)        p. 
0-7389  0-7935  100 

(1-3536  -  1)      (100-_rt 
0-7208  100      ' 

.-.  p  =  20-81.     (Observed,  20-71.) 

Problems  for  Solution 

PROBLEM  53. — At  17'4°  the  refractive  index  of  methyl 
alcohol  for  sodium  light  (WD)  =  1-3297  and  its  density  = 
0'7945.  Calculate  the  molecular  refractivity  according  to  both 
(a)  Gladstone  and  Dale's,  and  (b)  Lorentz  and  Lorenz's 
formula,  and  (c)  compare  the  value  for  the  latter  with  that 
calculated  from  the  atomic  refractivities. 

Ans.  (a)  13-27,  (b)  8-21,  (c)  8-23. 

PROBLEM  54. — d™'5  for  liquid  hydroxylamine  =  1-205  and 
WD  =  1-4405.  (a)  Calculate  the  molecular  refractivity  (L.  and 
L.).  (b)  What  is  the  atomic  refractivity  of  nitrogen  in  the 
compound  ? 

Ans.  (a)  7-23,  (b)  2-556. 

PROBLEM  55.— At  20°  the  density  of  chloroform  =  1-4823 
and  the  refractive  index  for  the  D  line  =  1-4472.  Given  the 
atomic  refractivities  of  carbon  and  hydrogen,  calculate  that  of 
chlorine  (L.  and  L.). 

Ans.  5-999. 


22  EEFEACTIVITY— PEOBLEMS 

PEOBLEM  56. — At  20°  the  substance  C3H6O  has  a  density  of 
0-8005  and  a  refractive  index  for  the  D  line  of  1-3641.  Using 
Lorentz  and  Lorenz's  formula,  determine  whether  it  has  the 
constitution  (a]  CH2  :  CH  •  CH2OH,  or  (b)  CH3  •  CO  •  CH3. 

Ans.  (b). 

PEOBLEM  57. — At  17 '5°  the  density  of  ethyl  alcohol  = 
0-8020,  and  its  refractive  index  for  the  D  line  =  1-3619.  At 
17 '8°  the  density  of  normal  propyl  alcohol  =  0*8074,  and  its 
refractive  index  for  the  D  line  =  1-3861.  From  these  clata 
calculate  the  molecular  refractivity  of  normal  butyl  alcohol 
according  to  Lorentz  and  Lorenz's  formula. 
Ans.  22-20,  (Observed,  22-11). 

PEOBLEM  58. — For  ethyl  acetate  d*6'4  =  0-9028  and  nD  = 
1*374:2.  (a)  Calculate  the  molecular  refractivity  (L.  and  L.), 
and  (b)  compare  it  with  the  value  calculated  from  the  atomic 
refractivities. 

Ans.  (a)  22-27,  (b)  22-38. 

PEOBLEM  59. — For  cinnamyl  alcohol 

C6H5.  CH  :  CH  .  CH2OH,  dj6'8  =  1-0555  and  nD  =  1-5763. 
(a)  Calculate  the  molecular  refractivity  (L.  and  L.),  and  (b) 
compare  it  with  the  value  calculated  from  the  atomic  re- 
fractivities. 

Ans.  (a)  42  03,  (b)  41-37. 

PEOBLEM  60. — At  the  temperature  t  the  density  of  water  is 
d*4  and  the  refractive  index  for  sodium  light  nD.  From  these 
data  test  the  constancy  of  the  specific  refractivity  as  calculated 
by  Gladstone  and  Dale's,  and  Lorentz  and  Lorenz's  formula. 

t  d\  n» 

20°  0-99823  1-3330 

40°  0-99224  1-3307 

80°  0-97183  1-3230 

PEOBLEM  61. — The  specific  refractivity  (L.  and  L.)  of 
glycerine  for  the  C  line  =  0-2219,  and  of  propyl  alcohol  = 
0-2903.  From  these  data  calculate  the  atomic  refractivity  of 
hydroxylic  oxygen  for  the  C  line. 

Ans.  1-50. 

PEOBLEM  62. — At  18-07°  the  density  of  ethylene  dibromide 
=  2-1830,  of  propyl  alcohol  =  0-80659,  and  of  a  mixture  of 
the  two  containing  p  per  cent,  by  weight  of  dibromide  = 
0*86081.  At  the  same  temperature  the  refractive  index  of 


EEFEACTIVITY— PROBLEMS  23 

the  dibromide  for  the  D  line  =  1-5404,  of  propyl  alcohol  = 
1-3862  and  of  the  mixture  =1-3919.     Calculate  the  value  of 

£,  (a)  using  the  Gladstone  and  Dale  formula,  (b)  using  the 
orentz  and  Lorenz  formula. 

Ans.  (a)  1017,  (b)  10-16.  (Observed,  lO'Ol.) 
PROBLEM  63. — At  18'07°  the  density  of  a  mixture  of  ethylene 
dibromide  and  propyl  alcohol  containing  20*95  per  cent,  of  the 
former  is  0-92908.  From  the  data  in  the  preceding  problem 
calculate  its  refractive  index  for  the  D  line  (a)  according  to 
Gladstone  and  Dale's  formula,  (b)  according  to  Lorentz  and 
Lorenz 's  formula. 

Ans.  (a)  1-39933,  (b)  1  39932.  (Observed,  1-39913.) 
PROBLEM  64.— At  20°  the  density  of  ether  =  0-72078,  of 
benzene  =  0*87953,  and  of  a  mixture  containing  p  per  cent, 
of  benzene  =  0*75299.  The  corresponding  refractive  indices 
are  (for  sodium  light),  ether  =  1-35360,  benzene  =  1-49996, 
mixture  =  I1 38227.  Calculate  the  value  of  p  by  the  Gladstone 
and  Dale  formula. 

Ans.  21-96.     (Observed,  21-06.) 

PROBLEM  65. — At  18-07°  the  density  of  a  5  per  cent,  sodium 
chloride  solution  =  1-0345  and  the  refractive  index  (D  line) 
=  1-3423.  At  the  same  temperature  the  density  of  water 
=  0-99866  and  the  refractive  index  =  1-3335.  Calculate  the 
specific  refractivity  of  sodium  chloride  (Gladstone  and  Dale). 
Ans.  0-274. 

PROBLEM  66. — At  18-07°  the  density  of  a_p  per  cent,  sodium 
chloride  solution  =  T0201  and  the  refractive  index  =  1-3388. 
Using  the  result  obtained  in  the  preceding  problem,  calculate 
the  value  of  p  (G.  and  D.). 

Ans.  3-005.     (Observed,  3-000). 

PROBLEM  67.— At  21-8°  the  density  ol  water  =  0-9978  and 
of  a  6-80  per  cent,  sodium  sulphate  solution  =  1-Q596.  At  the 
same  temperature  the  refractive  index  of  water  for  sodium 
:.ight  =  1-33308  and  of  the  solution  =  1-34291.  Calculate  the 
specific  refractivity  of  sodium  sulphate  by  the  Gladstone  and 
Dale  formula. 

Ans.  0-18382. 

PROBLEM  68. — At  21-8°  the  density  of  a  p  per  cent,  sodium 
sulphate  solution  =  1*0782  and  nD  =  1-34571.    From  the  data 
in  the  preceding  problem  calculate  the  value  of  p. 
Ans.  8-78.     (Observed,  8'80.) 


CHAPTBE  IV 

MOLECULAR  WEIGHT  FROM  LOWERING  OF  VAPOUR-PRES- 
SURE, LOWERING  OF  FREEZING-POINT  AND  ELEVATION 
OF  BOILING-POINT.—  DEGREE  OF  DISSOCIATION  OF  ELEC- 
TROLYTES.— MOLECULAR  LOWERING  OF  FREEZING-POINT 
AND  ELEVATION  OF  BOILING-POINT  FROMTLATENT  HEATS 
OF  FUSION  AND  EVAPORATION 

Molecular  Weight  from  Lowering  of  Vapour-pressure 

'"PHE  relative  lowering  of  the  vapour-pressure  of  a  pure 

JL      solvent  by  a  dissolved  substance  is  equal  to  the  ratio  of 

the  number  of  molecules  of  dissolved  substance  to  the  number 

of  molecules  of  solvent  in  the  dilute  solution  (Eaoult),  that  is, 

(,\  Po  ~  P  _  n 


where  p0  is  the  vapour-pressure  of  the  pure  solvent,  p  that  of 
the  solution  at  the  same  temperature,  n  the  number  of  mole- 
cules of  dissolved  substance,  and  N  the  number  of  molecules 
of  solvent  in  the  solution.  pQ  and  p  must,  of  course,  be 
measured  in  the  same  unit,  but,  since  we  are  here  dealing 
with  a  ratio,  the  unit  chosen  is  immaterial. 

This  relation  may  be  used  as  follows  for  the  determination 
of  the  molecular  weights  of  dissolved  substances.  If  a  is  the 
weight  of  the  dissolved  substance,  and  b  that  of  the  solvent 
in  the  solution,  m  the  molecular  weight  of  the  dissolved 
substance,  and  M  that  of  the  solvent  in  the  gaseous  state, 

then  the  number  of   molecules  of   dissolved  substance   =  — 

m 

and  the  number  of  molecules  of  solvent  =  -.      Therefore 


__      - 

N  ~  b/M  ~  bm 
24 


VAPOUR-PBESSUBE  AND  TEMPEBATUBE     25 

Vapour-pressure  and  Temperature 

The  relation  between  the  vapour-pressure  of  a  liquid  and 
the  temperature  is  expressed  by  the  equation  (Clausius), 


W       dT       ~  RT2' 

p  is  the  vapour-pressure  of  the  liquid  at  the  absolute  tempera- 
ture T,  L  the  molecular  heat  of  evaporation  of  the  liquid,  and 
'R  the  gas-constant,  the  numerical  value  of  which  is  1-985 
(in  round  numbers  2)  when  the  gram-calorie  is  the  unit  of 
energy. 

If  we  assume  that  L  is  independent  of  the  temperature  and 
integrate  equation  (3)  between  the  temperatures  T0  and  Tv 
we  obtain 

(4)  ^  a.  _£(!__  i)  _  L  (r'  -  T»\ 

where  p0  and  p1  are  the  vapour-pressures  corresponding  to 
the  absolute  temperatures  T0  and  Tlt  or,  converting  the  natural 
to  common  logarithms, 

(5)  log  ^=  log  p0  -  log  ^  = 

For  small  intervals  of  temperature  the  assumption  that  L 
is  constant  will,  in  most  cases,  be  practically  true.  For  larger 
intervals  of  temperature  L  may  be  taken  as  the  latent  heat 

T  +  T 
at  the  mean  temperature  -^-g — \      Since  in    equation   (5) 

we  are  concerned  only  with  the  ratio  of  the  vapour-pressures, 
pQ  and  p1  may  be  expressed  in  any  unit. 

Vapour-pressure — Examples 

PROBLEM  69. — What  is  the  concentration,  in  gram-mole- 
cules per  1000  grams  water,  of  an  aqueous  solution  which  at 
t0  =  100'42°  C.  has  a  vapour -pressure  p  =  758'2  mm.  of 
mercury?  The  molecular  heat  of  evaporation  of  water  is 
L  =  -  9600  cal. 

SOLUTION  69. — The  problem  can  be  solved  by  equation  (i) 
if  the  vapour-pressure  pQ  of  water  at  100-42°  C.  can  be  cal- 
culated. This  can  be  done  by  equation  (5)  as  follows : 
Under  a  pressure  p1  =  760  mm.  of  mercury  (atmospheric 
pressure)  water  boils  at  tl  =  100°  C.  or  Tl  =  373°  absolute, 
that  is,  at  T^  =  373  the  vapour-pressure  of  water  =  p  = 
760  mm. 


26  VAPOUE-PEESSUEE—  EXAMPLES 

T0  =  t0  +  273  =  100-42  +  273  =  373'42  and  L  =  -  9600. 
Substituting  these  numerical  values  in  (5),  we  obtain 


logp,  =  log  760  +  a.SxxsI^  -  2'89' 

•'•  Po  =  771  '2  mm. 

Accordingly,  by  (i), 

p0  -  p      771-2  -  758-2      Q-Q1685      _n 
~^~  771-2  ~1T~     =  ~F 

The  solution,  therefore,  contains  0-01685  gram-molecule  of 
dissolved  substance  per  gram-molecule  of  water,  and,  since 
the  gram-molecule  of  water  =  18  grams,  it  contains 

0-01685  x  1000 

—  -^  —      -  =  0*930    gram-molecule   of    dissolved   sub- 

stance per  1000  grams  water. 

PROBLEM  70.  —  Under  a  pressure  of  760  mm.  ether  boils 
at  ^  =  35°  C.  Its  molecular  heat  of  evaporation  is  -  6640 
calories,  (a)  What  is  the  vapour-pressure  p0  of  ether  at 
t0  =  30°  C.  ?  (b)  What  volume  will  be  occupied  by  a  litre 
of  air  at  a  pressure  P  =  720  mm.  after  being  bubbled  through 
ether  at  30°  C.,  if  the  total  pressure  be  kept  constant  at  720 
mm.  during  the  process,  and  (c)  how  many  (x)  grams  of 
ether  will  evaporate  during  the  process  ? 

SOLUTION  70.  —  (a)  The  vapour-pressure  at  30°  C.  may  be 
calculated  from  equation  (5).  T0  =  t0  +  273  =  303.  Tl  = 
$!  +  273  =  308.  p1  =  760.  L  =  -  6640.  Therefore 

-  6640  x  5 
log  Po  =  log  760  +  2-3  x  1-985  x  303  x  308 

and  p0  =  635  mm. 

(b)  The  volume  of  air,  which  has  been  bubbled  through 
ether  at  30°  C.  under  the  constant  pressure  P  =  720  mm., 
increases    until   the   partial   pressure    of    the    air    becomes 
p  =  P  -  pQt  where  _p0  is  the  vapour-pressure  of  ether  at  30°  C. 
From  Boyle's  law  we  get  the  final  volume  v  assumed  by  1 
litre  of  air  after  being  bubbled  through  the  ether.     Thus 

v_  _        P  720  720 

1  ~  P  -  p0  ~  720  -  635  "   85  ' 

.'.  v  =  8's  litres. 

(c)  These  8*5  litres  contain  ether  vapour  at  a  partial  pres- 
sure p0  =  635  mm.     The  weight  x  of  this  ether  vapour  is  to 
be  calculated. 


VAPOUR-PEESSUEE—  EXAMPLES  27 

According  to  equation  (2),  p.  1,  (see  also  Problem  2), 

Pov  =  ~ETQ1 
where  M  is  the  molecular  weight  of  ether  vapour.     Hence 


If  p0  and  v  are  expressed  in  atmospheres  and  litres  the  value 
of  R  is  0-08204:.     Therefore 


0-08204  x  303 

PBOBLEM  71. — The  vapour-pressure  of  a  solution*  con- 
taining 6-69  grams  of  Ca(N03)2  in  100  grams  of  water  is 
746-9  mm.  at  100°  C.  What  is  the  degree  of  dissociation  of 
the  salt  ? 

SOLUTION  71. — Let  a  be  the  degree  of  dissociation,  then,  since 
Ca(N03)2  dissociates  into  3  ions,  one  Ca"  and  two  NO'3,  one 
gram-molecule  of  Ca(N03)2  gives  on  dissociation  a  gram-mole- 
cule of  Ca"  and  2a  gram-molecule  of  N0'3,  whilst  (1  -  a) 
gram-molecule  of  undissociated  salt  remains.  The  total 
number  of  molecules  derived  from  one  gram-molecule  of 
Ca(N03)2  is,  therefore,  1  -  a  +  3a  =  1  +  2a.  From  6-69 

>  *  A  •  (\  O 

grams  =  -v«irT  gram-m°lecules  °f  Ca(N03)2  there  are,  there- 
fore, derived 

fi'fiQ 

n  =  jgjTj  (1  +  2a)  gram-molecules, 

164-1  being  the  molecular  weight  of  Ca(N03)2. 
According  to  (i) 

pQ  -  p  =  n_m 
Po       ~  N' 

p  =  746-9  mm.  and  p0  for  pure  water  at   100°  =  760  mm. 
N  =  weight  of  water  in  the  solution  divided  by  the  molecular 

100 

weight  of  water  as  vapour  =    — — .     Hence 

18 

760  -  746-9  =  6-69   +  ^  _  18 
760      164*1 

/.  a  =  0-675. 


28  LOWERING  OF  FREEZING-POINT 

Lowering  of  Freezing-point 

If  t°  is  the  freezing-point  of  a  pure  solvent,  and  t^  that  of 
a  solution  containing  c  gram- molecules  of  dissolved  substance 
per  1000  grams  of  solvent,  the  lowering  of  the  freezing- 
point  A  =  t  -  tj,  is  proportional  to  c,  that  is,  the  lowering  of 
the  freezing-point  of  a  pure  solvent  by  a  dissolved  substance 
is  proportional  to  the  concentration  of  the  latter,  or 

(6)  A  =  Kc. 

The  value  of  the  constant  K  depends  only  on  the  solvent, 
and  on  the  unit  of  concentration  chosen.  We  shall  take  as 
our  unit  concentration,  one  gram-molecule  of  solute  in  1000 
grams  of  solvent.  If  in  (6)  we  put  c  =  1,  we  obtain  A  =  K. 
K  is,  therefore,  the  lowering  of  the  freezing-point  of  the 
solvent  caused  by  dissolving  1  gram-molecule  of  a  normal 
(non-dissociating  and  non-associating)  solute  in  1000  grams 
of  solvent.  It  is  called  the  molecular  lowering  of  the  freezing- 
point. 

(Two  other  units  of  concentration  are  often  used,  namely, 

(1)  one  gram-molecule  of  solute  in  100  grams  of  solvent,  and 

(2)  one  gram-molecule  of  solute  in  1  gram  of  solvent.     If  K' 
is  the  molecular  lowering  in  the  first  case,  and  K"  that  in  the 
second  case,  the  relation  between  K,  K  and  K'  is  K'  =  10  K, 
K"  =  1000  K.) 

If  the  solution  contains  a  grams  of  solute  in  b  grams  of 
solvent,  and  if  M  is  the  molecular  weight  of  the  solute,  the 

CL 

concentration  of  the  solute  is  -r?  gram-molecules  in  b  grams 

of  solvent,  or     ^ —  gram-molecules  per  1000  grams  solvent. 

Therefore,  from  (6), 

,  N  ^  1000  a 

(7)  A-*-E5- 

By  various  transformations  of  this  equation  any  of  the 
quantities  may  be  calculated  if  the  others  are  known.  In 

(?)>   — £ —  is  the  weight  of  solute  in  1000  grams  of  solvent. 
If  we  put  this  =  W,  the  equation  becomes 

W        A        W 
(8)  A  =  tf-^or-^  =^, 

a  form  which  is  particularly  clear  and  easy  of  application. 


FREEZING-POINT—EXAMPLES  29 

The  constant  K  for  a  given  solvent  may  be  found  empiric- 
ally by  (6),  (7)  or  (8),  by  observing  the  depressions  of  the 
freezing-point  of  the  pure  solvent  caused  by  known  concen- 
trations of  substances  of  which  the  molecular  weights  are 
known,  and  taking  the  mean  value  ;  or  it  may  be  calculated 
from  the  latent  heat  of  fusion  of  the  solvent  by  means  of  the 
relation  (van't  Hoff) 


T  (  =  273  +  t}  is  the  freezing-point  of  the  pure  solvent  on  the 
absolute  scale,  I  the  latent  heat  of  fusion  of  1  gram  of  the 
solvent  in  calories  and  R  the  gas-constant  (1*985,  or,  in  round 
numbers,  2). 

(If  the  constants  K'  or  K"  as  denned  above  are  used,  then 

RT*  RT* 


Elevation  of  Boiling-point 

For  the  elevation  qf  the  boiling-point  of  a  pure  solvent  by 
a  dissolved  substance  equations  exactly  analogous  to  (6),  (7), 
(8)  and  (9)  hold.  In  this  case,  however,  A  is  the  elevation 
of  the  boiling-point  and  is  equal  to  ^  -  £,  where  t  is  the 
boiling-point  of  the  pure  solvent  and  tl  that  of  a  solution  con- 
taining c  gram-molecules  of  solute  of  molecular  weight  M  per 
1000  grams  of  solvent,  or  a  grams  of  solute  in  b  grams  of 
solvent,  or  W  grams  of  solute  in  1000  grams  of  solvent.  K 
is  the  molecular  elevation  of  the  boiling-point,  caused  by 
dissolving  one  gram-molecule  of  a  normal  solute  in  1000 
grams  of  solvent.  In  equation  (9)  T  (=  273  +  t)  is  the 
boiling-point  of  the  pure  solvent  on  the  absolute  scale  and  I 
is  the  latent  heat  of  evaporation  of  one  gram  of  solvent  in 
calories. 

(As  in  the  case  of  the  freezing-point,  K'  an.d  K"  are  the 
molecular  elevations  of  the  boiling-point  when  either  of  the 
other  common  units  of  concentration  are  used.  Their 
relations  to  K  and  I  are  thesame  as  those  given  under  the 
freezing-point.) 

Freezing-point  —  Examples 

PROBLEM  72.  —  The  freezing-point  of  a  solution  of  0-684 
gram  of  cane  sugar  in  100  grams  of  water  is  -  0'037°  C.,. 


28  LOWERING  OF  FREEZING-POINT 

Lowering  of  Freezing-point 

If  t°  is  the  freezing-point  of  a  pure  solvent,  and  t^  that  of 
a  solution  containing  c  gram- molecules  of  dissolved  substance 
per  1000  grams  of  solvent,  the  lowering  of  the  freezing- 
point  A  =  t  —  £x  is  proportional  to  c,  that  is,  the  lowering  of 
the  freezing-point  of  a  pure  solvent  by  a  dissolved  substance 
is  proportional  to  the  concentration  of  the  latter,  or 

(6)  A  =  Kc. 

The  value  of  the  constant  K  depends  only  on  the  solvent, 
and  on  the  unit  of  concentration  chosen.  We  shall  take  as 
our  unit  concentration,  one  gram -molecule  of  solute  in  1000 
grams  of  solvent.  If  in  (6)  we  put  c  =  1,  we  obtain  A  =  K. 
K  is,  therefore,  the  lowering  of  the  freezing-point  of  the 
solvent  caused  by  dissolving  1  gram-molecule  of  a  normal 
(non-dissociating  and  non-associating)  solute  in  1000  grams 
of  solvent.  It  is  called  the  molecular  lowering  of  the  freezing- 
point. 

(Two  other  units  of  concentration  are  often  used,  namely, 

(1)  one  gram- molecule  of  solute  in  100  grams  of  solvent,  and 

(2)  one  gram-molecule  of  solute  in  1  gram  of  solvent.     If  K 
is  the  molecular  lowering  in  the  first  case,  and  K"  that  in  the 
second  case,  the  relation  between  K,  K'  and  K"  is  K'  =  10  K, 
K"  =  1000  K.) 

If  the  solution  contains  a  grams  of  solute  in  b  grams  of 
solvent,  and  if  M  is  the  molecular  weight  of  the  solute,  the 

concentration  of  the  solute  is  -^  gram-molecules  in  b  grams 

of  solvent,  or     ^, —  gram-molecules  per  1000  grams  solvent. 

Therefore,  from  (6), 

„  1000  a 


By   various   transformations   of  this  equation  any  of  the 
quantities  may  be  calculated  if  the  others  are  known.      In 

(?)>   — £ —  is  tne  weight  of  solute  in  1000  grams  of  solvent. 
If  we  put  this  =  W,  the  equation  becomes 

(«J~A-*-»<>'t-ir.  1 

a  form  which  is  particularly  clear  and  easy  of  application. 


FKEEZING-POINT—  EXAMPLES  29 

The  constant  K  for  a  given  solvent  may  be  found  empiric- 
ally by  (6),  (7)  or  (8),  by  observing  the  depressions  of  the 
freezing-point  of  the  pure  solvent  caused  by  known  concen- 
trations of  substances  of  which  the  molecular  weights  are 
known,  and  taking  the  mean  value  ;  or  it  may  be  calculated 
from  the  latent  heat  of  fusion  of  the  solvent  by  means  of  the 
relation  (van't  Hoff) 


T  (  =  273  +  f)  is  the  freezing-point  of  the  pure  solvent  on  the 
absolute  scale,  I  the  latent  heat  of  fusion  of  1  gram  of  the 
solvent  in  calories  and  R  the  gas-constant  (1*985,  or,  in  round 
numbers,  2). 

(If  the  constants  K  or  K"  as  denned  above  are  used,  then 

RT*  RT* 


Elevation  of  Boiling-point 

For  the  elevation  Oyf  the  boiling-point  of  a  pure  solvent  by 
a  dissolved  substance  equations  exactly  analogous  to  (6),  (7), 
(8)  and  (p)  hold.  In  this  case,  however,  A  is  the  elevation 
of  the  boiling-point  and  is  equal  to  ^  -  t,  where  t  is  the 
boiling-point  of  the  pure  solvent  and  ^  that  of  a  solution  con- 
taining c  gram-molecules  of  solute  of  molecular  weight  M  per 
1000  grams  of  solvent,  or  a  grams  of  solute  in  b  grams  of 
solvent,  or  W  grams  of  solute  in  1000  grams  of  solvent.  K 
is  the  molecular  elevation  of  the  boiling-point,  caused  by 
dissolving  one  gram-molecule  of  a  normal  solute  in  1000 
grams  of  solvent.  In  equation  (p)  T  (=  273  +  t)  is  the 
boiling-point  of  the  pure  solvent  on  the  absolute  scale  and  I 
is  the  latent  heat  of  evaporation  of  one  gram  of  solvent  in 
calories. 

(As  in  the  case  of  the  freezing-point,  K  an4  K"  are  the 
molecular  elevations  of  the  boiling-point  when  either  of  the 
other  common  units  of  concentration  are  used.  Their 
relations  to  K  and  /  are  thesame  as  those  given  under  the 
freezing-point.) 

Freezing-point  —  Examples 

PROBLEM  72.  —  The  freezing-point  of  a  solution  of  0-684 
gram  of  cane  sugar  in  100  grams  of  water  is  -  0'037°  C.,. 


32  VAPOUE-PEESSUEE— PEOBLEMS 

128  x  2-1  _ 

128        ~  2l' 
Substituting  this  value  of  K  in  (1)  we  obtain 

1-985  x  (308)2 
1  -     1000  x  2-1      =  8<>'7  calories- 

Since  the  molecular  weight  of  ether  is  74,  the  molecular 
heat  of  evaporation  is 

L  =  89-7  x  74  =  6638  calories. 

PBOBLEM  75. — A  solution  of  barium  nitrate  containing 
a  =  11-07  grams  in  100  grams  of  water  boils  at  100-466°  C. 
What  is  the  degree  of  dissociation  of  the  salt  ?  K  for  water 
=  0-52. 

SOLUTION  75. — The  solution  contains  10  a  grams  barium 
nitrate  per  1000  grams  water.  If  M  (  =  261-5)  is  the  mole- 
cular weight  of  barium  nitrate,  it  would,  therefore,  contain 

-jg-  gram-molecules  in  1000  grams  water  if  the  salt  were 

undissociated.  Let  a  be  the  degree  of  dissociation.  Since 
Ba(NO3)2  dissociates  into  3  ions  we  obtain,  exactly  as  in 
Problem  71,  for  the  total  number  of  gram-molecules  in  1000 
grams  of  water 

10  a  , 

C  =   -^-(1  +  2  a). 

Putting  this  value  of  c  into  equation  (6)  we  obtain 

A  -*.^-(l  +  2  a), 

and,  substituting  the  numerical  values, 

..._       0-52  x  10  x  11-07  „  • 

261-5         '   (1  +  2a) 
/.  a  =  0-558. 

Problems  for  Solution 

Vapour-pressure 

PROBLEM  76. — The  vapour-pressure  of  ether  at  20°  C.  is 
442  mm.  and  that  of  a  solution  of  6'1  grams  of  benzpic  acid 
in  50  grams  of  ether  is  410  mm.  at  the  same  temperature. 
Calculate  the  molecular  weight  of  benzoic  acid  in  ether. 
Ans.  124. 


VAPOUR-PRESSURE—PROBLEMS  33 

PROBLEM  77. — At  10°  C.  the  vapour-pressure  of  ether  is     , 
291-8  mm.  and  that  of  a  solution  containing  5'3  grams  ot   / 
benzaldehyde  in  50  grams  of  ether  is  £71  '8  mm.     What  is^ 
the  molecular  weight  of  benzaldehyde  ? 
Ans.  1144. 

PROBLEM  78. — The  vapour-pressure  of  alcohol  at  70°  C.  is        ^ 
54:0-9  mm.,  and  at  80°  C.  it  is  811-8  mm.    Calculate  the  latent 
heat  of  evaporation  per  gram  of  alcohol.  \J 

Ans. — 212  calories. 

PROBLEM  79. — Under  a  pressure  of  760  mm.  ether  boils       / 
at  35°  C.     A  solution  of  10'44  grams  of  aniline  in  100  grams      / 
of  ether  has  a  vapour-pressure   of  333   mm.   at    15'3°   C.      ( 
The  latent  heat  of  evaporation  of  ether  is  —  89'73  calories 
per  gram.     Calculate  the  molecular  weight  of  aniline  in  the 
solution. 

Ana.  97-6. 

PROBLEM  80. — At  0°  C.  the  vapour- pressure  of  water  is 
4-620  mm.  and  of  a  solution  of  8'49  grams  of  NaN03  in  100 
grams  of  water  4*483  mm.     Calculate  the  degree  of  dissocia-\/ 
tion  of  the  NaNO3. 

Ans.  0-649. 

PROBLEM  81. — At  25°  C.  the  vapour-pressure  of  water  is 
23-55  mm.  What  is  the  vapour-pressure  of  a  solution  con- 
taining  6  grams  of  urea  in  100  grams  of  water  atf'the  same 
temperature  ? 

Ans.  23-13  mm.  / 

PROBLEM  82. — At  0°  C.  the  vapour-pressure  of  Water  = 
4"620  mm.  and  that  of  a  solution  containing  21'24ygrams  of 
glycerol  in  100  grams  of  water  =  4-432  mm.  ^Calculate  the 
molecular  weight  of  glycerol  in  the  solution. 

Ans.  93-9. 

PROBLEM  83. — The  vapour-pressure  of  a  solution  of  8-89 
grams  of  dextrose  in  100  grams  of  water  is  4-532  mm.  at 
0°  C.,  whilst  that  of  pure  water  at  the  same /temperature 
is  4-620  mm.  Calculate  the  molecular  weigfit  of  dextrose. 

Ans.  194-5. 

PROBLEM  84. — A  solution  containing  9'21  grams  of  mercuric 
cyanide  in  100  grams  of  water  has  a  vapour-pressure  of 
755-2  mm.  at  100°  C.  What  is  the  molecular  weight  of  the 


34  VAPOUR-PRESSURE—PROBLEMS 

7 

salt?     What  conclusion  may  be  drawias/to  the  dissociation 
of  mercuric  cyanide  in  aqueous  solutions 
Ans.  262-5. 

PROBLEM  85. — The  vapour-pressure  of  a  solution  con- 
taining 11-94  grams  of  glycocoll  (C.,H5NO2)  iti  lOCXgrams  of 
water  is  740-9  mm.  at  100°  C.  What  is  the  mole/ular  weight 
of  the  solute  ? 

Ans.  85-5.  / 

PROBLEM  86. — Under  what  pressure  does /water  boil  at  a 
temperature  of  95°  C.  ?  The  latent  heat  oft  evaporation  of 
water  is  -  536  calories  per  gram. 

Ans.  636-5  mm. 

PROBLEM  87. — At  0°  C.  the  vapour-pressure/of  water  = 
4-620  mm.  and  that  of  a  solution  of  2-21  grams  of  CaCl2  in 
100  grams  of  water  =  4'585  mm.  CaWla^e  the  apparent 
molecular  weight  and  the  degree  of  dissockCtion  of  the  CaCl2. 

Ans.  M  =  52-5,  a  =  0-557. 

PROBLEM  88. — A  current  of  dry  air  was  bubbled  through  a 
bulb  containing  a  solution  of  13-33  grams  of  urea  in  100  grams 
of  water,  then  through  a  bulb,  at  the  same  temperature,  con- 
taining pure  water  and  finally  through  &  tmje  containing 
pumice  moistened  with  strong  sulphuric  acid.  The  loss  of 
weight  of  the  water  bulb  =  0'0870  grams  and  the  gain  of 
weight  of  the  sulphuric  acid  tube  =  2-036  grams.  Calculate 
the  molecular  weight  of  urea  in  the  solution. 

Ans.  56-2. 

PROBLEM  89. — A  current  of  dry  air  was  passed  first  through 
a  series  of  bulbs  containing  a  solution  of  8-914  grams  of 
nitrobenzene  in  100  grams  of  alcohol,  and  therr  through  a 
series  of  bulbs  containing  pure  alcohol.  Th£  temperature  was 
11°  C.  After  the  passage  of  the  air  the  decrease  in  the  weight 
of  the  bulbs  containing  the  solution  was  2^340  grams  and 
of  the  bulbs  containing  the  pure  solvent  0-0685  gram.  Cal- 
culate the  molecular  weight  of  nitrobenzene  in  the  solution. 
Ans.  125-8. 

PROBLEM  90  (of.  preceding  problem). — In/a  similar  ex- 
periment with  ethyl  benzoate  the  solution  contained  7'394 
grams  of  ethyl  benzoate  in  100  grams  of  "aj/ohol.  The  loss 
of  weight  of  the  solution  was  2-1585  grams  and  of  the  pure 


FKEEZING-POINT— PKOBLEMS  35 

solvent   Q'0515   gram.     Calculate  the   molecular   weight  of 
ethyl  benzoate. 

Ans.  146. 

PROBLEM  91. — The  vapour-pressure  of  boron  trichloride  is 
562-9  mm.  at  10°  C.  and  807'5  mm.  at  20°  C.  )Vhat  is  the 
molecular  heat  of  evaporation  of  boron  trichloride  and  its 
boiling-point  under  atmospheric  pressure  ? 

Ans.  L  =  -  5936  cals.,  B.P.=  184° C. 

Freezing-point 

PROBLEM  92. — The  freezing-point  of  pure  benzene  =  5-440° 
and  that  of  a  solution  containing  2-093  grams  of  benzalde- 
hyde  in  100  grams  of  benzene  =  4-440°.  Calculate  the  mole- 
cular weight  of  benzaldehyde  in  the  solution.  K  for  benzene 
=  5. 

Ans.  104-6. 

PROBLEM  93. — A  solution  of  0'502  gram  of  acetone  in 
100  grams  of  glacial  acetic  acid  gave  a  depression  of  the 
freezing-point  of  0-339°.  Calculate  the  molecular  depression 
for  glacial  acetic  acid. 

Ans.  K  =  3-9. 

PROBLEM  94. — 17*79  grams  of  an  aqueous  solution  contain- 
ing 0-1834  gram  of  hydrogen  peroxide  gave  a  freezing-point 
of  -  0'571°.      What  is  the  molecular   weight   of  hydrogen 
peroxide  in  the  solution  ?     K  for  water  =  1-86. 
Ans.  33-9. 

PROBLEM  95. — By  dissolving  0'0821  gram  of  w-hydroxy- 
benzaldehyde  (C7H6O2)  in  20  grams  of  naphthalene  (melting- 
point  80- 1°)  the  freezing- point  is  lowered  by  0'232°.  Assum- 
ing that  the  molecular  weight  of  the  solute  is  normal  in 
the  solution,  calculate  the  molecular  depression  for  naphtha- 
lene and  the  latent  heat  of  fusion  per  gram. 

Ans.  K  =  6-896,  I  =  36-2  cals. 

PROBLEM  96. — A  solution  of  1  gram  of  silver  nitrate  in 
50  grams  of  water  freezes  at  -  0-348°  C.  Calculate  to  what 
extent  the  salt  is  ionised  in  the  solution.  K  for  water  =  T86. 

Ans.  a  =  0'59.  v 

PROBLEM  97. — The  weights  a  of  methyl  alcohol  dissolved 


36  FBEEZING-POINT— PKOBLEMS 

in  15  grams  of  benzene  gave  the  depressions  of  the  freezing- 
point  A. 

a  0-0478    0-0988    0-2700    0-4291     0*6636     1-093 
A  0-360°     0-612°     1-265°     1-610°     1-978°     2-475°. 

The  molecular  lowering  of  the  freezing-point  for  benzene  is 
K  =  5-0.  What  conclusions  as  to  the  molecular  condition  of 
methyl  alcohol  in  benzene  solution  may  be  drawn  from  these 
figures  ? 

PROBLEM  98. — An  aqueous  solution  of  ethyl  alcohol  con- 
taining 8-74  grams  alcohol  per  1000  grams  water  gave  a 
freezing-point  of    -0'354°.      Find  the  molecular  weight  of 
alcohol  in  this  solution.     K  for  water  =  1*86. 
Ans.  45-9. 

PROBLEM  99. — A  solution  of  NaCl  containing  3-668  grams 
per  1000  grams  water  freezes  at   -  0-2207°.     Calculate  the 
degree  of  dissociation  of  the  salt.     (K  —  1*86.) 
Ans.  a  =  0-892. 

PROBLEM  100. — 0-2274  gram  of  naphthalene  in  10  grams 
of  _p-toluidine  (melting-point  =  42'1°)  caused  a  lowering  of  the 
freezing-point  =  0'940°.      Calculate   the  freezing-point  con- 
stant and  latent  heat  of  fusion  of  j>toluidine. 
Ans.  K  =  5-29,  I  =  37'5  cals. 

PROBLEM  101. — The  lowering  of  the  freezing-point  A  was 
caused  by  dissolving  a  grams  of  formanilide  in  10  grams  of 
benzene.  (K  =  5'0.) 

a        0-1255        0-3815        0*7439        1-197 
A        0-420          0-9'20          1-360          1-747 

What  conclusions  as  to  the  state  of  the  dissolved  substance 
and  its  variation  with  concentration  may  be  drawn  from  these 
figures  ? 

PROBLEM  102.— A  solution  of  0-4180  gram  KOH  in  1000 
grams  water  gave  a  freezing-point  of  -  0-0275°,  and  a  solution 
of  1*780  gram  in  1000  grams  water  a  freezing-point  of 
-  0-1147°.  Calculate  the  degree  of  dissociation  of  the  KOH 
in  these  solutions,  taking  K  =  1-86. 

Ans.  98-1  per  cent,  and  94  per  cent. 

PROBLEM  103. — The  melting-point  of  phenol  is  40°  C. 
A  solution  containing  0-172  gram  acetanilide  (C8H9ON)  in 
12*54  grams  phenol  freezes  at  39'25°  C.  Assuming  that 


FREEZING-POINT—PROBLEMS  37 

acetanilide  has  its  normal  molecular  weight  in  phenol,  cal- 
culate the  freezing-point  constant  and  the  latent  heat  of 
fusion  of  phenol. 

Ans.  K  =  7-38,  I  =  26'5  cals. 

PROBLEM  104. — A  solution  of  8*535  grams  NaNO3  in  100 
grams  water  freezes  at  -  3 '04:°  C.     Calculate  the  degree  of 
dissociation  of  the  NaN03.     K  =  1-86. 
Ans.  0-629. 

PROBLEM  105. — The  freezing-point  of  a  solution  containing 
0-510  gram-molecule  of  strontium  formate  in  1000  grams 
water  is  —  2-390°.  Calculate  the  degree  of  dissociation  of 
the  salt.  K  =  1-86. 

Ans.  0-76. 

PROBLEM  106. — The  freezing-point  of  a  solution  of  barium 
hydroxide  containing  1  gram-molecule  in  64  litres  is  -  0'0833°. 
What  is  the  concentration  of  hydroxyl-ions  in  the  solution  ? 
Take  K  =  1*89  for  concentrations  in  gram-molecules  per 
litre. 

Ans.  0-0284  gram-ion  per  litre. 

PROBLEM  107. — A  solution  containing  2-423  grams  sulphur 
in  100  grams  naphthalene  (melting-point  =  80-1°)  gave  a 
lowering  of  the  freezing-point  of  0!641°,  and  a  solution  con- 
taining 2-192  grams  iodine  in  100  grams  naphthalene  a  depres- 
sion of  0*595°.  The  latent  heat  of  fusion  of  naphthalene  is 
35-7  calories  per  gram.  What  is  the  molecular  formula  of 
sulphur  and  iodine  respectively  in  naphthalene  solution  ? 
Ans.  M  for  S  =  264,  .-.  S8,  M  for  I 

PROBLEM  108. — A  solution  containing  0'063 
of  calcium  formate  in  1000  grams  water  fr< 
What  is  the  degree  of  dissociation  ?     K  =  1- 
Ans.  0-85. 

PROBLEM  109. — A  solution  containing  J0K)2  'gram-molecule 
of  zinc  chloride  per  litre  freezes  at  -\0>tu35°.     Calculate  the 
degree  of  dissociation.     K  =  1-89. 
Ans.  0-87. 

PROBLEM  110. — A  solution  of  zinc  nitrate  containing  0-065 
gram-molecule  per  litre  freezes  at    -  0-322°.     What  is  the 
concentration  of  the  zinc  ions  in  the  solution?     K  =  1-89. 
Ans.  0-0526  gram-ion  per  litre. 


38  FEBBZING-POINT— PROBLEMS 

PROBLEM  111. — The  melting-point  of  pure  copper  is 
1084°  C.  A  solution  of  Cu2O  in  copper,  containing  1-16  per 
cent,  by  weight  of  Cu2O  freezes  at  1076°  C.  Assuming  that 
the  molecular  weight  of  Cu2O  in  the  solution  corresponds  to 
its  formula,  calculate  the  latent  heat  of  fusion  of  copper  per 
gram. 

Ans.  38  cals. 

PROBLEM  112. — A  solution  of  lithium  chloride  containing 
4-13  grams  per  litre  freezes  at  -  0-343°.  What  is  the  degree 
of  dissociation?  K  =  1-89. 

Ans.  0-865. 

PROBLEM  113.— The  melting-point  of  tin  =  231-611°  C. 
and  its  latent  heat  of  fusion  =  14*25  calories  per  gram.  The 
freezing-point  of  a  solution  containing  1*5463  grams  copper 
in  440  grams  tin  is  229*692°.  Calculate  the  molecular  weight 
of  copper  in  the  solution. 

Ans.  65-4. 

PROBLEM  114. — The  freezing-point  of  a  solution  containing 
0*0199  gram-molecule  SrCl2  per  litre  is  -  0-1015°.     What  is 
the  degree  of  dissociation  of  the  salt?     K  =  1*89. 
Ans.  0*85. 

PROBLEM  115. — A  solution  containing  0*834  gram  Na2S04 
per  1000  grams  water  freezes  at  -  0'0280°.  Assuming  dissocia- 
tion into  3  ions,  calculate  the  degree  of  dissociation  and  the 
concentrations  of  the  Na-  and  SO4"  ions.  K  =  1*86. 

Ans.  a  =  0-782  ;  cone.  Na-  =  0*0918  gram-ion  per  litre  ;  cone. 
S04"  =  0*0459  gram-ion  per  litre. 

PROBLEM  116. — The  vapour-pressure  of  water  at  0°  C.  is 

4*620  mm.,  and  the  depression  of  the  vapour-pressure  caused 

by  dissolving  5*64  grams  NaCl  in  100  grams  water  is  0*142 

mm.     What  is  the  freezing-point  of  this  solution?   /K  =  1*86. 

Ans.  -  3-177°.  / 

PROBLEM  117. — The  vapour-pressure  of  a  solution  con- 
taining 5*85  grams  NaCl  in  100  grams  water  /s  4-460  mm. 
at  0°  C.,  and  that  of  pure  water  is  4*620  mmXyThe  freezing- 
point  of  the  solution  is  -  3*424°.  Compare  the  degrees  of 
dissociation  obtained  (1)  by  the  vapour-pressure  method,  and 
(2)  by  the  freezing-point  method.  K  =  1-86. 

Ans.  (1)  0*924,  (2)  0-841. 


BOILING-POINT— PROBLEMS  39 

PROBLEM    118. — When    mercuric  cyanide,    Hg(CN)2,    is 
dissolved  in  potassium  cyanide  solutions,  the  complex  anion 
Hg(CN)m  +  2  is  formed  according  to  the  equation 
Hg(GN)2  +  mCN'  =  Hg(CN)m  +  2. 

The  freezing-point  of  a  solution  containing  0*1965  gram- 
molecule  of  KCN  per  litre  is  -  0-704°,  and  that  of  the  same 
solution,  after  the  addition  of  0'095  gram-molecule  of 
Hg(CN)2  per  litre,  is  -  0'530°.  What  is  the  value  of  m? 
K  =  1-86.  As  m  must  be  a  whole  number,  give  the  nearest 
whole  number. 

Ans.  m  =  1-985,  .-.  2. 

PROBLEM  119. — The  freezing-point  of  a  solution  containing 
0-01052  gram-molecule  Na2Si03  in  1000  grams  water  is 
-  0'0676°.  Show  that  the  salt  is  largely  hydrolysed  accord- 
ing to  the  equation 

Na2Si03  +  H2O  =  2  NaOH  +  SiO2  (colloidal). 

PROBLEM  120. — The  freezing-point  of  a  0*25  N-KCN  solu- 
tion is  -  0-860°.  The  freezing-point  of  the  same  solution 
with  the  addition  of  0-25  gram-molecule  of  AgCN  per  litre 
is  -  0-830°.  The  solution  of  AgCN  in  KCN  takes  place 
according  to  the  equation 

m  ON'  +  AgCN  =  (AgCN)  (CN')m. 

What  is  the  value  of  m  (nearest  whole  number)  ?     K  =  1-86. 
Ans.  m  =  1-06,  /.  1. 

Boiling-point 


PROBLEM  121. — A  solution  containing  0-5042  gram  of  a 
substance  dissolved  in  42-02  grams  of  benzene  boils  at  80-175°. 
Find  the  molecular  weight  of  the  s6lute,  having  given  that 
the  boiling-point  of  benzene  is  80.000°,  and  its/ latent  heat  of 
evaporation  of  94  calories  per  gram. 
Ans.  181-9. 


PROBLEM  122. — A  solution  containing  0-7269  gram  cam- 
phor (mol.  wt.  =  152)  in  32-08  £ram<  of  acetone   (boiling- 
point  =  56-30°  C.)  boiled  at  56-55°TT    What  is  the  molecular 
elevation  for  acetone  and  the  latent  heat  of  evaporation? 
Ans.  K  =  1-674,  I  =  129*5  cals.  per  gram. 

PROBLEM  123. — The  latent  heat  of  evaporation  of  carbon 
disulphide  (boiling-point  =  46- 20°  C.)  is  85*9  cals.  per  gram. 


40  BOILING-POINT—  PROBLEMS 

The  weights  a  of  benzole  acid  dissolved  in  50-09  grams  of  CS2 
gave  the  elevations  A  of  the  boiling-point  of  CS2.  What  is 
the  molecular  condition  of  benzoic  acid  in  CS2,  and  how  does 
it  vary  with  concentration  ? 

a  0-9378       1-6429      2-5792       4-5519  grafns. 
A  0187°       0-319°       0-479°       0-789°.  / 

PBOBLEM  124.  —  A  solution  of  9*472  gram/  CdI2  in  44-69 
grams  water  boiled  at  100'303°.  The  latent  heat  of  evapora- 
tion of  water  is  536  cals.  per  gram.  What  is  the  molecular 
weight  of  CdI2  in  the  solution  ?  ^(haj/conclusion  as  to  the 
state  of  CdI2  in  solution  may  be  drawn  from  the  result  ? 
Ans.  M  =  363-2. 

PBOBLEM  125.  —  The  boiling-point  of  a  solution  of  0-4388 
gram  NaCl  in  100  grams  water  is  100-074°  C.  /Calculate  the 
apparent  molecular  weight  of  the  NaCl  and  its  degree  of 
dissociation.  K  =  0-52.  \  / 

Ans.  M  =  30-84,  a  =  \g&7. 

PROBLEM  126.  —  The  boiling-point  of  acetic  acid  is  118-100° 
and  its  latent  heat  of  evaporation  121  cals.  per  gram.  A 
solution  containing  0*4344  gram  of  anthracene/m  44-16  grams 
of  acetic  acid  boils  at  118-240°.  What  is  the  nlolecular  weight 
of  anthracene?  \J 

Ans.  178. 

PROBLEM  127.  —  By  dissolving  3  '614  grams  of  CuCl2  in  100 
grams  of  alcohol  the  boiling-point  of  the  alcohol  is  raised 
0-308°.  For  alcohol  K  =  1-15.  What  is  the  molecular  weight 
of  the  solute  ? 

Ans.  134-9. 

.    PROBLEM  128.  —  The    boiling-point   of   a  solution  of  3'40 
grams  BaCl2  in  100   grams  water  ts  100-208°.     K  =  0'52. 
What  is  the  degree  of  dis:ociation  ofV^BaCl^? 
Ans.  0-725. 

PROBLEM  129.  —  At  100°  the  vapour  pressure  of  a  solution 
of  6-48  grams  NH4C1  in  100  grams  water  is  731-4  mm.  K  = 
0*52.     What  is  the  boiling-point  of  the  solution? 
Ans.  101-086°. 

.?/  - 


CHAPTER  V 

SUKFACE  TENSION,   MOLECULAR  WEIGHT  AND  DEGBEE  OF 
ASSOCIATION  OF  LIQUIDS 

Definitions— Example 

PROBLEM  130.— In  a  capillary  tube  of  radius  r  =  0  01425 
JL  cm.,  pure  liquid  formic  acid  rises  to  a  height  of  h^  = 
4-442  cms.  at  ^  =  16-8°,  hz  =  4-205  cms.  at  t2  =  46-4°  and 
hs  -  3-90  cms.  at  tz  =  79'8°.  The  density -of  formic  acid  is 
dl  =  1-207  at  16-8°,  d2  =  1-170  at  46-4°  and  d3  =  1-129  at 
79*8°.  What  is  the  molecular  weight  of  formic  acid  in  the 
liquid  state,  and  how  does  it  vary  with  temperature? 
(g  =  981-1  cms./sec.2). 

SOLUTION  130. — The  surface  tension  of  a  liquid  of  density 
d,  as  determined  by  the  height  h  to  which  it  rises  in  a 
capillary  tube  of  radius  r,  is 

(i)  7  =  I  grhd, 

where  g  is  the  value  of  gravity  (gravitation  constant).     If  the 
quantities   on  the  right-hand   side  of  the  equation  are   ex- 
pressed in  C.G.S.  units,  g  in  cms./SQc.2,  r  and    h   in  cms. 
and  d  in  grams  per  c.c.,  we  obtain  y  in  dynes  per  cm. 
If  d  is  the  density  of  the  liquid  under  investigation,  and 

M  its  molecular  weight,  its  specific  volume  is  v  =  -3  and  its 

molecular  volume  is  Mv.  The  molecular  surface  of  the 
liquid  is,  therefore,  proportional  to  (Mv)*l*.  '  The  product 
y  (Mv)2l*  is  called  the  molecular  surface  energy,  and,  if  we 
use  the  above  units,  is  expressed  in  ergs.  The  molecular 
surface  energy  of  any  liquid  diminishes  with  rise  of  tempera- 
ture, and  Ramsay  and  Shields  have  shown  that  the  tempera- 
ture coefficient  of  the  molecular  surface  energy  is  the  same 
for  all  liquids,  and  that  its  numerical  value  is  2-121  when  the 
surface  energy  is  expressed  in  ergs.  Hence 

41 


42  SUEPACE  TENSION—  EXAMPLES 


where  y15  i^  are  the  surface  tension  and  specific  volume  at 
the  lower  temperature  ^  and  yv  v2  the  corresponding  values 
at  the  higher  temperature  £2. 

The  molecular  weight  of  the  substance  in  the  liquid  state 
is  not,  however,  necessarily  the  same  as  the  normal  mole- 
cular weight,  or  formula  weight,  M.  This  is  the  case  with 
unassociated  liquids  only.  If  the  molecules  in  the  liquid 
condition  are  associated  into  more  or  less  complex  groups, 
we  must,  in  order  to  make  equation  (2)  generally  applicable, 
multiply  the  normal  molecular  weight  M  by  a  factor  x,  so 
that  xM  expresses  the  mean  molecular  weight  of  the  substance 
in  the  liquid  state  between  the  temperatures  ^  and  t2.  x  is 
called  the  factor  of  association,  and  in  the  case  of  unassociated 
liquids  its  value  is  unity/  Equation  (2),  therefore,  becomes 


For  formic  acid  at  ^  =  16-8°  we  have 

7i  =  i  fl^Mi  =  °'5  x  981'1  x  °'01425  x  4'442  x  1-207 
=  37'47  dynes  per  cm., 

at  £2  =  464° 

y2  =  4  9rhA  =  °'5  x  981<1  x  0*01425  x  4-205  x  1-170 

=  34'42  dynes  per  cm., 
at  £3  =  79-8° 

7s  =  i  grhA  =  °'5  x  981'1  x  0-01425  x  3-90  x  1-129 

=  30*80  dynes  per  cm. 
Therefore, 

y^Mvrfl*  =  37'47(46/l-207)2/8  =  424-4  ergs 
2/3  =  34-42(46/l-170)2/3  =  397'7    „ 
=  30-80(46/1-129)2/3  =  364-6    „ 

Putting  these  values  into  equation  (3)  we  obtain,  between  ^ 
and  J2 

^(424-4  -  397-7) 

46-4  -  16-8  21> 


2-121  x  29-6 

26^7 

/.  x  -  (2-352)  3/2  =  3-60. 


,,          -        x      - 

26^7  =  ' 


SUEFACE  TENSION^-PKOBLEMS  43 

The  mean  molecular  weight  of  formic  acid  between  16'8°  and 
46-4°  is,  therefore,  3'60  x  46  =  166. 
Similarly,  between  tiJL  and  t3, 

0^(397-7-364-6) 

79-8  -  46-4  "' 

.-.  x  =  3-13, 

and  the  mean  molecular  weight  of  formic  acid  between  46-4° 
and  79-8°  is  3-13  x  46  =  144. 

The  molecules  of  formic  acid  are,  therefore,  associated  into 
complex  groups  in  the  liquid  state,  and  the  degree  of  associa- 
tion, and,  therefore,  the  molecular  weight  of  the  complexes  dim- 
inish with  increase  of  temperature. 

Problems  for  Solution 

PROBLEM  131. — What  is  the  factor  of  association  of  carbon 
disulphide,  for  which  at  19-4°  the  surface  tension  is  33 '58 
dynes  per  cm.,  and  the  density  1*264,  and  at  46 '1°  the  surface 
tension  is  29-41  dynes  per  cm.,  and  the  density  1'223? 
Ans.  1-07. 

PROBLEM  132. — Liquid  nitrogen  peroxide  rises  to  a  height 
of  3*14  cms.  at  1-6°  in  a  capillary  tube  of  0*0129  cm.  radius, 
and  to  a  height  of  2-905  cms.  in  the  same  tube  at  19*8°  0. 
The  density  of  the  peroxide  is  1-486  at  1-6°,  and  1-444  at 
19'8°.  From  these  data  determine  whether  liquid  nitrogen 
peroxide  is  associated  at  these  temperatures,  and  if  so,  to 
what  extent. 

Ans.  x  for  N204  =  1*01  /.  formula  is  N2O4. 

PROBLEM  133. — Normal  butyl  alcohol  at  the  temperature 
t°  rose  to  a  height  h  in  a  capillary  tube  of  radius  0*01425 
cms.  The  density  of  the  alcohol  at  each  temperature  was  d. 
Calculate  the  factor  of  association  and  the  mean  molecular 
weight  between  each  pair  of  temperatures. 


t    17-4° 

45-7° 

77-9°  0. 

h    4-305 

4-005 

3-63  cms. 

d   0-8115 

0-7907 

0-7634. 

Ans.  x 

1-94 

1-72 

xM 

143-5 

127-3 

PROBLEM  134. — In  a  capillary  tube  of  0-03686  cms.  di- 
ameter water  rose  at  0°  C.  to  a  height  of  8-10  cms.,  and  at 


44  SUEFACE  TENSION— PEOBLEMS 

10°  C.  to  a  height  of  7 '96  cms.  The  density  of  water  at 
0°  =  0-9999  and  at  10°  =  0-9997.  What  is  the  factor  of  as- 
sociation and  the  mean  molecular  weight  of  liquid  water 
between  0°  and  10°  ? 

Ans.  x  =  3-81,  xM  =  68-6. 

PROBLEM  135. — In  a  capillary  tube  of  radius  0'0129  cms. 
ethyl  iodide  rose  to  the  height  h  at  the  temperature  t,  at 
which  the  density  is  d, 

t  h  d 

19-1°  2-445  cms.  1'937 

46-2°  2-22  cms.  1-875. 

What  is  the  factor  of  association  and  mean  molecular  weight 
of  ethyl  iodide  between  these  two  temperatures  ? 

Ans.  x  =  I'Ol.     Molecular  weight  .-.  normal. 


CHAPTER  VI 
THEBMOCHEMISTEY 

Mess's  Law 

r  I  "HE  heat  effect  of  a  chemical  reaction,  for  a  given  quantity 
J-  of  the  reacting  substances  and  for  a  definite  temperature, 
is  dependent  only  on  the  initial  and  final  states  of  the  reacting 
system.  It  is  the  same  whether  the  reaction  takes  place  in 
one  or  in  several  stages,  provided  that  the  initial  and  final 
states  are  the  same  (Hess's  Law).  By  means  of  this  law  we 
are  enabled  to  calculate  the  heat  effect  of  a  reaction  which, 
for  various  reasons,  cannot  be  measured  directly.  The  re- 
action in  question  is  considered  as  the  sum  or  difference  of 
others  which  have  be"en  measured.  By  suitable  manipulation 
of  the  thermochemical  equations  all  undesired  substances  are 
eliminated,  and  only  the  equation  for  the  required  reaction  is 
left. 

By  the  union  of  12  grams  of  solid  carbon  with  32  grams  of 
gaseous  oxygen  at  18°  C.  to  form  44  grams  of  gaseous  carbon 
dioxide,  there  is  a  heat  evolution  of  96960  calories,  or,  the 
energy-content  of  the  system  in  its  initial  state  (12  grams 
carbon  +  32  grams  oxygen)  is  greater  than  the  energy-content 
of  the  system  in  its  final  state  (44  grams  carbon  dioxide)  by 
96960  calories.  This  statement  is  conveniently  expressed 
by  the  equation 

C  +  02  =  CO2  +  96960  cals. 

The  symbols  in  such  thermochemical  equations  stand  for 
definite,  though  unknown,  amounts  of  energy  associated  with 
the  formula-weight  in  grams  of  each  substance.  A  knowledge 
of  the  absolute  amount  of  energy  associated  with  each  gram- 
molecule  is,  however,  unnecessary,  since  it  is  only  with 
differences  in  energy  that  we  are  concerned. 

45 


46  THEKMOCHEMISTEY 

Heat  of  Reaction  at  Constant  Volume  and  Constant 
Pressure 

The  difference  in  energy-content  between  the  initial  and 
final  states  of  the  system  is  equal  to  the  heat-evolution  or 
absorption  only  when  no  external  work  is  done  during  the 
reaction.  This  is  the  case  when  only  solids  and  liquids  are 
involved  in  the  reaction,  since  in  such  cases  the  volume 
changes  during  the  reaction  are  negligible.  When  gases  are 
involved  in  the  reaction,  however,  the  heat  of  the  reaction  at 
constant  pressure  may  differ  considerably  from  that  at  con- 
stant volume,  since,  in  the  formation  or  absorption  of  one 
gram- molecule  of  a  gas  at  T°  absolute,  external  work  equiva- 
lent to  ET  =  %T  calories  is  done  by  or  on  the  system,  and 
the  total  energy- difference  between  the  initial  and  final  states 
of  the  system  is  equal  to  the  heat  evolved  +  the  external 
work  done  by  the  system. 

If  Qv  is  the  heat  of  the  reaction  in  calories  at  constant 
volume  (no  external  work  condition),  and  Qp  that  at  constant 
pressure  (external  work  condition),  then 

Qv  =  Qp  +  nET 

=  Qp  +  ZnT  calories, 

where  n  is  the  number  of  gas-molecules  in  the  final  state  (on 
the  right-hand  side  of  the  equation  representing  the  reaction) 
in  excess  of  the  number  in  the  initial  state  (on  the  left-hand 
side  of  the  equation),  and  Tthe  absolute  temperature  at  which 
the  reaction  takes  place. 

The  heat  of  formation  of  a  compound  is  the  heat  evolved 
in  the  formation  of  a  gram-molecule  of  the  compound  from 
its  component  elements. 

As  may  be  readily  seen  from  Hess's  law,  the  heat  evolved 
in  a  chemical  reaction  is  equal  to  the  sum  of  the  heats  of 
formation  of  the  final  substances  minus  the  sum  of  the  heats 
of  formation  of  the  original  substances,  the  heats  of  formation 
of  the  elements  themselves  being  taken  as  zero. 

In  the  following  problems  the  symbol  Aq  denotes  a  large 
quantity  of  water.  For  example, 

NaCl  +  Aq  =  NaCl  Aq 

means  the  solution  of  1  gram-molecule  of  NaCl  in  much 
water  to  form  a  dilute  solution. 


I .  t  THBBMOGHBMISTBY— EXAMPLES  47 

Thermochemistry— Examples 

PEOBLEM  136. — The  heat  of  combustion  of  ethyl  alcohol 
is  Ql  =  341800  cals.  ;  the  heats  of  formation  of  carbon  di- 
oxide and  water  are  Q>2  =  96000  cals.  and  Q3  =  68000  cals. 
respectively,  all  at  constant  pressure.  What  is  the  heat  of 
formation  Qx  of  ethyl  alcohol  ? 

SOLUTION  136. — The  combustion  of  ethyl  alcohol  takes 
place  according  to  the  equation 

(1)  C2H5OH  +  302  =  2C02  +  3H,0  +  Ql  cals. 

The  combustion  can  be  regarded  as  taking  place  in  3  stages, 
(a)  in  the  decomposition  of  the  alcohol  into  its  elements  ac- 
cording to  the  equation 

(2)  C2H5OH  =  20  +  3H2  +  O  -  Qx  cals., 

and  (b)  and  (c)  in  the  combustion  of  20  and  3H2  according  to 
the  equations 

(3)  2C  +  202  =  2C02  +  2g2 
and 

(4)  3H2  +  liO2  =  3H20  +  3g3. 

By  adding  (2),  (3)  and  (4)  we  obtain 

C2H6OH  +  302  -  2C02  +  3H20—  Qx  +  2Q2  +  SQy- 
From  a  comparison  of  this  equation  with  (1)  it  follows  that 

Ql  =  -  Qx  +  2Q2  +  3g3, 
and,  therefore, 

Qx  =  2<?2  +  3Q3  -  Q1 

=  192000  +  204000  -  341800  =  54200  cals. 

to  V^c*^ 

PEOBLEM  137. — By  the  solution  of  a  =  10  grams  of  metallic 
sodium  in  much  water  q1  =<18800  cals.  are  liberated,  and  by 
the  solution  of  b  =  20  grams  of  sodium  oxide  under  the  same 
conditions  q%  ='  20400  cals.  are  liberated.  What  is  the  mole- 
cular heat  of  formation  Qx  of  Na2O,  if  the  molecular  heat  of 
formation  of  liquid  water  from  gaseous  oxygen  and  hydrogen 
is  Q3  =  68000  cals.  ? 

SOLUTION  137. — The  solution  of  1  gram-atom  of  metallic 
Na  in  water  takes  place  according  to  the  thermochemical 
equation 

(1)  Na  +  H,0  =  NaOH  +  |H2  +  Qlt 


48  THERMOCHEMISTRY— EXAMPLES 

and  the  solution  of  1  gram-molecule  of  Na2O  according  to  the 
equation 

(2)  Na2O  +  H2O  =  2NaOH  +  C2. 

By  multiplying  equation  (1)  by  2  and  subtracting  equation  (2 
we  obtain 

(3)  2Na  +  H20  =  Na2O  +  Ha  +  2QX  -  Qz. 
To  obtain  the  heat  of  formation  of  Na2O  we  must  add  to 
equation  (3)  the  equation  representing  the  formation  of   a 
gram-molecule  of  liquid  water  from  gaseous  hydrogen  and 
oxygen,  namely  : — 

(4)  H2  +  i02  =  H20  +  Q8. 
From  (3)  and  (4)  we  thus  get 

2Na  +  iOa  =  Na2O  +  2QX  -  Q2  +  Qy 
The  required  heat  of  formation  of  Na.20  is  therefore 
(5)  &=2gi  +  Q3-  Q2. 

If  M  is  the  atomic  weight  of  Na  and  Mr  the  molecular  weight 
of  Na2O,  then 

Ql  -  2£  and  Q2  -  &JL'f 
or,  putting  in  the  numerical  values, 


e 

- 


and 

.     o   =  20400  x  62 

20 
Accordingly,  from  (5), 

Qx  =  2  x  43200  +  68000  -  63200  =  91200  cals. 

PROBLEM  138. — Auric  hydroxide  dissolves  in  hydrochloric 
acid  according  to  the  thermochemical  equation 

Au(OH)3  +  4HC1  =  HAu014  +  3H2O  +  23000  cals, 

and  in  hydrobromic  acid  according  to  the  corresponding 
equation 

Au(OH)3  +  4HBr  =  HAuBr4  +  3H2O  +  36800  cals. 

On  mixing  I  gram-molecule  HAuBr4  with  4  gram-molecules 
HC1  there  is  a  heat  absorption  of  510  cals.  What  percentage 
of  the  bromoauric  acid  has  been  transformed  into  chloroauric 
acid  in  the  process  ? 


THERMOCHEMISTEY— PROBLEMS  49 

SOLUTION  138. — The  ratio  of  the  quantity  x  of  bromoauric 
acid  transformed  into  chloroauric  acid  to  the  total  quantity  of 
bromoauric  acid  employed  is  equal  to  the  ratio  of  the  observed 
heat  evolution  to  the  molecular  heat  of  the  reaction 

HAuBr4  +  4HC1  =  HAuCl4  +  4HBr. 

The  molecular  heat  of  this  reaction  is  obtained  by  subtracting 
equation  (2)  from  equation  (1),  giving  equation  (3). 

(1)  Au(OH)3  +  4HC1  =  HAuCl4  +  3H2O  +  23000  cals. 

(2)  Au(OH)3  +  4HBr  =  HAuBr4  +  3H9O  +  36800  cals. 

(3)  HAuBr4  +  4HC1  =  HAuCl4  +  4HBr  -  13800  cals. 

For  the  quantity  x  (per  cent.)  transformed  we  therefore  get 

jc_         -  510 
100        -  13800' 
x  =  3*7  per  cent. 

Problems  for  Solution 

PROBLEM  139. — At  17°  C.  the  heat  of  combustion  of  carbon 
to  carbon  dioxide  is  96960  cals.,  and  that  of  carbon  monoxide 
to  carbon  dioxide  67960  cals.,  both  at  constant  pressure.- 
What  is  the  heat  of  formation  of  carbon  monoxide  (a)  at 
constant  pressure,  (b)  at  constant  volume  ? 

Ans.  (a)  29000  cals.,  (b)  29290  cals. 

PROBLEM  140. — By  the  combustion  at  constant  pressure  of 
2  grams  of  hydrogen  with  oxygen  to  form  liquid  water  at  17°  C. 
68360  cals.  are  evolved.  What  is  the  heat  evolution  at  con- 
stant volume? 

Ans.  67490  cals. 

PROBLEM  141. — The  heat  of  solution  of  MgSO4  is  20280 
cals.,  of  MgSO4,H2O,  13300  cals.,  and  of  MgSO4, 7H2O, 
-  3800  cals.  What  is  the  heat  of  hydration 

(a)  of  MgSO4  to  MgS04,  H2O, 

(b)  of  MgS04  to  MgS04l  7H2O, 

(c)  of  MgS04,  H2O  to  MgSO4,  7H2O  ? 

Ans.  (a)  6980  cals.,  (b)  24080  cals.,  (c)  17100  cals. 

PROBLEM  142. — From  the  following  data  calculate  the  heat 
of  formation  of  potassium  hydroxide : — 
4 


50  THERMOCHEMTSTKY— PEOBLEMS 

K  +  H2O  +  Aq  =  KOH  Aq  +  H  +  48100  cals. 

H2  +  O  -  H2O  +  68400  cals. 
KOH  +  Aq  =  KOH  Aq  +  13300  cals. 
Ans.  103200  cals. 

PROBLEM  143. — The  heat  of  solution  of  BaCl0  is  2070  cals., 
and  its  heat  of  hydration  to  BaCl2,  2H2O  is  6970  cals.     What 
is  the  heat  of  solution  of  the  latter  salt  ? 
Ans.    -  4900  cals. 

PROBLEM  144. — At  ordinary  temperature  the  heats  of  com- 
bustion of  12  grams  of  diamond,  graphite  and  amorphous 
carbon  are  94310,  94810  and  97650  cals.  respectively.  What 
is  the  heat  of  formation  (a)  of  diamond  from  amorphous 
carbon,  (b)  of  graphite  from  amorphous  carbon,  (c)  of  diamond 
from  graphite  at  ordinary  temperature  ? 

Ans.  (a)  3340,  (b)  2840,  (c)  500  cals. 

PROBLEM  145. — From  the  following  data  calculate  the  heat 
of  formation  of  anhydrous  A12C16  : — 

2A1  +  6HC1  Aq  =  Al2Cl6Aq  +  3H2  +  239760  cals. 

H2  +  Clj,  =  2HC1  +  44000  cals. 

HC1  +  Aq  =  HC1  Aq  +  17315  cals. 

A12C16  +  Aq  =  Al2Cl6Aq  +  153690  cals. 

Ans.  321960  cals. 
PROBLEM  146. — The  heats  of  solution  of 

Na2SO4  ;  Na2SO4,  H2O  :  and  Na2SO4,  10H2O 

are  460,  -  1900  and  -  18760  cals.  respectively.  What  are 
the  heats  of  hydration  of  Na2SO4  (a)  to  monohydrate,  (b)  to 
decahydrate  ? 

Ans.  (a)  2360,  (b)  19220  cals. 

PROBLEM  147. — The  heat  of  formation  of  H20  is  68360 
cals.  and  of  CO2,  96960  cals.,  both  at  17°  C.  and  constant 
pressure.  The  heat  of  combustion  of  methane  at  17°  C.  and 
constant  pressure  is  211930  cals.  Calculate  the  heat  of  for- 
mation of  methane  at  17°  (a)  at  constant  pressure,  (b)  at 
constant  volume. 

Ans.  (a)  21750  cals.,  (b)  21170  cals. 

PROBLEM  148. — The  heat  of  neutralisation  of  HC1  with 
NaOH  is  13680  cals.,  that  of  acetic  acid  with  the  same  base 
is  13400  cals.,  and  of  butyric  acid  13800  cals.  What  are  the 


THERMOCHEMISTRY— PROBLEMS  51 

heats  of  dissociation  (a)  of  acetic,  (b)  of  butyric  acid,  if  both 
are  regarded  as  practically  undissociated  ? 

Ans.  (a)  -  280,  (b)  120  cals. 

PROBLEM  149. — The  heats  of  neutralisation  of  NaOH  and 
NH4OH  by  HC1  are  13680  and  12270  cals.  respectively. 
What  is  the  heat  of  dissociation  of  NH4OH,  if  it  is  assumed 
to  be  practically  undissociated  ? 

Ans.    -  1410  cals. 

PROBLEM  150. — The  heat  of  neutralisation  of  HNO3  by 
NaOH  is  13680  cals.  and  of  CHC12 .  COOH,  14830  cals. 
When  one  equivalent  of  NaOH  is  added  to  a  dilute  solution 
containing  one  equivalent  of  HNO3  and  one  equivalent  of 
CHC12 .  COOH,  13960  cals.  are  liberated.  In  what  ratio  is 
the  base  distributed  between  the  two  acids  ? 

Ans.  HNO3 :  CHC12 .  COOH  =  0-756 :  0-244  =  3-1 : 1. 

PROBLEM  151. — The  heats  of  formation  of  C02,  liquid  H,,O 

and  C2H4  at  17°  and  constant  pressure  are  96960,  68360  and 

-  2710  cals.  respectively.     What  is  the  heat  of  combustion 

of  C2H4  at  17°  to  C02  and  liquid  H2O  (a)  at  constant  pressure, 

(b)  at  constant  volume  ? 

Ans.  (a)  333350  cals.,  (b)  332190  cals. 

PROBLEM  152. — From  the  following  data  calculate  the  heat 
of  formation  of  HNO2  Aq. 

NH4N02  =  N2  +  2H20  +  71770  cals. 
2H2  +  O2  =  2H20  +  136720  cals. 
No  +  3H2  +  Aq  =  2NH3Aq  +  40640  cals. 
NHjAq  +  HN0.2Aq  =  NH4NO2Aq  +  9110  cals. 
NH4N02  +  Aq  =  NH4NO2Aq  -  4750  cals. 

Ans.  H  +  N  +  02  +  Aq  =  HNO2Aq  +  30770  cals. 

PROBLEM  153. — The  heat  of  neutralisation  pf  hydrochloric 
acid  by  sodium  hydroxide  is  13780  cals.,  and  of  monochlor- 
acetic  acid  14280  cals.  When  one  equivalent  of  hydrochloric 
acid  is  added  to  one  equivalent  of  sodium  monochloracetate 
in  dilute  aqueous  solution  there  is  a  heat  absorption  of  455 
cals.  How  much  of  the  acetate  is  decomposed  according  to 
the  equation 

CH2C1 .  COONa  +  HC1  =  NaCl  +  CH2C1.  COOH  ? 
Ans.  0'91  equivalent. 


52  THERMOCHEMISTRY— PROBLEMS 

PROBLEM  154. — From  the  following  data  : — 

C  +  O2  =  CO2  +  96960  cals., 
2H2  +  O2  =  2H.O  +  136720  cals., 
2CfiH6  +  15O2  =  12CO2  +  6H20  +  1598Wcals.. 
2C2H2  +  502  =  4CO2  +  2H2O  +  620100^18., 

all  at  17°  and  constant  pressure,  calculate  the  heat  evolved  at 
17°  in  the  reaction 

3C2H2  =  C6H6, 

(a)  at  constant  pressure,  (b)  at  constant  volun^e, 

Ans.  (a)  130800  cals.,  (b)  129640  cals. 

PROBLEM  155. — From  the  following  data  calculate  the  heat 
of  formation  of  As2O3  : — 

As2O3  +  3H2O  +  Aq  =  2H3As03Aq  -  7550  calsY 

As  +  301  =  As013  +  71390  cals.v- 
AsCL  +  3H2O  +  Aq  =  H3AsO3Aq  +  3HC1  Aq  +  17580  cals. 

H  +  01  =  HC1  +  22000  cals. 
HC1  +  Aq  =  HOI  Aq  +  17315  cals. 
H2  +  O  =  H2O  +  68360 

Ans.  154680  cals. 


CHAPTER  VII 
VELOCITY  OF  KEACTION 

Monomolecular  Reaction 

THE  equation  for  the  velocity  at  any  instant  of  a  mono- 
molecular  reaction  which  goes  to  practical  completion 
is,  at  constant  temperature, 


a  is  the  initial  concentration  of  the  substance  undergoing 
change  and  x  the  amount  changed  after  the  time-interval  t 
from  the  commencement  of  the  reaction,  dx  represents  the 
infinitely  small  quantity  of  substance  transformed  in  the 
infinitely  small  time-interval  dt,  starting  from  t.  x  is  not,  of 
course,  the  absolute  amount  changed  but  is  the  decrease  in 
the  initial  concentration  due  to  the  change,  k  is  called  the 
velocity-constant  of  the  reaction. 
This  equation  gives,  on  integration, 

(1)    -  loge  (a  —  x)  =  kt  +  constant. 
But  when  t  =  0,  x  =  0, 

(2)  .*.  -  logea  =  constant. 
By  subtracting  (2)  from  (1)  we  obtain 

loge     a      =  kt  or  k  =  -  loge      a     , 
&  a  -  x  t     5  e  a  -.x 

or,  converting  to  common  logarithms, 


a  -  x 


If  xl  and  x2  are  the  amounts  transformed  after  the  time- 
intervals  2j  and  &2  respectively,  then  from  (2), 

2-302  log  —  -  —  =  to.  and  2-302  log 

° 


a  -  xl  a  -  x2 

53 


54  VELOCITY  OF  KEACTION 

and,  by  subtracting  the  former  equation  from  the  latter, 


or> 


log 


In  the  case  of  a  monomolecular  reaction  the  numerical  value 
of  k  is  independent  of  the  unit  chosen  to  express  the  concen- 
trations a  and  x.  Thus  if  a  unit  n  times  less  than  that  used 
in  equation  (2)  is  chosen,  we  obtain 


log  -  =  log  __  =  *. 

*  w(a  -  #)  £  a  -  a; 

Bimolecular  Reaction 

At  constant  temperature  the  velocity  at  any  instant  of  a 
bimolecular  reaction  which  goes  to  practical  completion  is 
given  by  the  equation 

(4)  *?  -  k(a  -  x)(b  -  x). 

a  and  b  are  the  initial  concentrations  of  the  reacting  sub- 
stances, x  the  diminution  in  concentration  of  the  reacting 
substances  after  the  time-interval  t  from  the  commencement 
of  the  reaction,  due  to  their  transformation  into  the  products 
of  the  reaction,  and  k  the  velocity-constant. 
Equation  (4)  gives,  on  integration, 

*  *  (a  -  b)t  10ge  (6  -  x)a 

,       _     2-302    .      (a  -  x)b 

~  (a  -  b)t     g(6  -  x)a 

When  the  initial  concentrations  of  the  reacting  substances 
are  the  same,  i.e.  when  a  =  6,  equation  (4)  becomes 


which  gives,  on  integration, 

(6)*-l  * 


t     a(a  -  x) 

For  a  bimolecular  reaction  the  numerical  value  of  k  depends 
on  the  unit  of  concentration  chosen  for  a,  b  and  x.     Thus  if 


VELOCITY  OF  EEACTION— EXAMPLES    55 

a  unit  I/nth  of  that  used  in  equations  (5)  and  (6)  is  employed, 
these  equations  become 

2-302     ,      n(a  -  x)bn  2-302    ,      (a  -  x}b 

f—\    ]f'  __  lOg  — =   1O2  J '— 

w  n(a  -  b)t    &  n(b  -  x)an      n(a  -  b)t        (b  -  x)a 

and(8)it'  =  i.          m  *-•'• 


t     na(a  -  x)n      t     na(a  -  x)' 
By  dividing  (7)  by  (5)  or  (8)  by  (6)  we  obtain 


Therefore,  when  using  a  particular  value  of  k  for  a  given 
reaction,  the  concentrations  should  be  expressed  in  the  same 
unit  as  was  employed  in  obtaining  that  value  of  k. 

Velocity  of  Reaction — Examples 

PROBLEM  156. — When  a  solution  of  dibromsuccinic  acid  is 
heated  the  acid  decomposes  into  brom-maleic  acid  and  hydro- 
bromic  acid  according  to  the  equation 

CHBr  .  COOH      CH  .  COOH 
|  =||  +  HBr. 

CHBr .  COOH      CBr  .  COOH. 

At  50°  the  initial  titre  of  a  definite  volume  of  the  solution  was 
T0  =  10-095  c.c.  of  standard  alkali.  After  t  minutes  the  titre 
of  the  same  volume  of  solution  was  Tt  c.c.  of  standard  alkali. 

*          0  214  380 

Tt      10-095  (T0)       10-37         10-57 

(a)  Calculate  the  velocity-constant  of  the  reaction,  (b)  After 
what  time  is  ^  of  the  dibromsuccinic  acid  decomposed  ? 

SOLUTION  156. — (a)  The  initial  equivalent  concentration  a 
of  the  dibromsuccinic  acid  is  proportional  to  T0.  Since  1 
molecule  of  dibasic  dibromsuccinic  acid  gives.  1  molecule  of 
dibasic  brom-maleic  acid  and  1  molecule  of  monobasic  hydro- 
bromic  acid,  the  increase  in  the  titre  (Tt  -  T0)  after  time  t 
is  proportional  to  the  equivalent  concentration  of  the  hydro- 
bromic  acid  at  time  t.  But  the  formation  of  1  equivalent  of 
hydro bromic  acid  means  the  disappearance  of  2  equivalents 
of  dibromsuccinic  acid,  therefore  x,  the  decrease  in  the  equi- 
valent concentration  of  the  dibromsuccinic  acid  after  time  t, 
is  proportional  to  2  (Tt  -  T0).  Therefore,  from  (2), 


56        VELOCITY  OF  EEACTION— EXAMPLES 

,       2-302  .          a          2-302  ,_  T0 

k  =  _— —  log - 


i 

For  t  =«  214  minutes 


•*»!*„£ 


and  for  £  *  380  minutes 


.  x  UH  x  10*7 

The  mean  velocity-constant  is,  therefore,  k  =  0*000260. 

(b)  When  ^  of  the  acid  is  decomposed  the  concentration  of 
the  remaining  acid  is  f  of  the  initial  concentration  a  ;  there- 
fore (a  -  x)  =  fa  and 


See  also  problem  234,  p.  108. 

PROBLEM  157.  —  A  b  =  O'Ol  N-solution  of  ethyl  acetate  is 
saponified  to  the  extent  of  c  =  10  per  cent,  in  ^  =  23  minutes 
by  an  a  =  0*002  N-solution  of  sodium  hydroxide.  In  how 
many,  (£2),  minutes  would  it  be  saponified  to  the  same  extent 
by  an  a'  =  0'005  N-solution  of  potassium  hydroxide  ? 

SOLUTION  157.—  The  saponification  of  ethyl  acetate  takes 
place  according  to  the  equation 

CH3  .  COOC2H5  +  H.20  =  CH3  .  COOH  +  C2H5OH. 

In  alkaline  solutions  free  acetic  acid  is  not  formed,  but  the 
alkali  salt,  which  is  almost  completely  dissociated  into  its 
ions.  The  equation  for  the  reaction  may,  therefore,  be  written 
in  the  ionic  form 

CH3  .  COOC2H5  +  OH'  =  CH3  .  COO'  +  C2H5OH.  • 
If  x  is  the  quantity  of  ethyl  acetate  saponified  at  the  time  t, 
the  velocity  of  saponification  —  is  given  by  the  equation 


.  COOC2H5], 

k(a  -  x)(b  -  x), 


VELOCITY  OF  EEACTION—  EXAMPLES         57 

if  the  NaOH  is  regarded  as  completely  dissociated.  The 
square  brackets  denote  the  concentration  of  the  enclosed  sub- 
stance. The  integration  of  this  differential  equation  is  carried 
out  as  follows  :  — 

7  -  m  -  ,  -  *•* 

(a  -  x)(b  -  x) 


(a  -  0)  \o  -  x       a  -  x 
and,  on  integration, 

(1)    -  —-1  [log«  (b  ~  x)  ~  loge(«  -  &)]  =  kt  +  constant. 


At  the  time  t  =  0,  x  =  0,  and,  therefore, 

(2)        —  —  [loge&  -  logea] 
By  subtracting  (2)  from  (1)  we  obtain 


(2)        —  —  [loge&  -  logea]  =  constant. 


and,  converting  to  common  logarithms, 

(3)    *-  ±?P.2,  .log*£jl*J. 

t(a  -  b)     *  a(b  -  x) 

From  the  data  gjven  in  tLs  problem,  at  the  time  ^  =  23 
minutes  x  =  -^-6  =  —b  =  O'l  b.    The  velocity-  constant  k 

may,  therefore,  be  calculated  from  equation  (3),  since  all  the 
other  quantities  are  known. 

For  this  particular  case  we  have 

k  =  ' 


^(a  -  b)         a(b  -  O'l    ) 
2-302        l       0-01  x  0-001 


23  x  -  0-008     °  0-002  x  0-009 
=  -  12-51  log  0-556 
=  -  12-51  x    -  0-255  =  3-19. 

It  is  now  possible  to  calculate  the  time  t%  in  which  the  same 
ester  solution  would  be  saponified  to  the  same  extent  by 
a'  =  0-005  N-KOH  solution.  Since  KOH,  like  NaOH, 
may  be  regarded  as  completely  dissociated  in  dilute  solution, 
we  obtain  again 

=  k(a'  -  x)(b  -  x), 


58         VELOCITY  OF  EEACTION—  EXAMPLES 
or,  integrated, 


Tc(a'  -  b)       ea'(b  -  x)' 

2-302  Jo     0-01  x  0-004 


3-19  x    -  0-005     6  0-005  x  0-009' 
=»  -  144  log  0-89  =  144  x  0'051  =  7-34  minutes. 

PROBLEM  158  (cf.  preceding  problem). — What  do  the  times 
$!  and  £2  in  the  preceding  problem  become  (1)  if  all  the 

concentrations  a,  b  and  a'  are  diminished  to  -  =  —  of  their 

n       10 

values  there,  (2)  if  the  temperature  is  raised  by  15°  C..  if  it  is 
assumed  that  the  velocity-constant  for  any  temperature  is 
doubled  by  a  rise  of  10°  G.  ? 

SOLUTION  158. — (1)  If  all  the  concentrations  in  the  preced- 
ing problem  are  diminished  to  -  ~  TQ  of  their  values  there, 
the  new  values  are 

a  =  0-0002,  b  =  0-001,  a'  =  0-0005,  x  =  0-0001. 
As  in  the  preceding  problem  we  obtain 

t,  =     2'302     log  b(a  ~  *)' 
1      k(a  -  b)     g  a(b  =!3f 

2-302  ^     0-001  x  0-0001 


.      3-19  x   -  0-0008     &  0-0002  x  0-0009' 

=  -  900  log  0-556  =  230  minutes, 
and  similarly, 

tj  =   -  1440  log  0-89  =  73-4  minutes. 

If,  therefore,  all  the  concentrations  are  diminished  to  -  of 

n 

their  previous  values,  the  time  required  for  the  reaction  to  pro- 
ceed to  the  same  extent  is  increased  to  n  times  its  previous  value. 
(2)  If  a  rise  of  temperature  of  10°  C.  doubles  the  velocity- 
constant  k,  i.e.  if 

^t°  + 10  =  2fcto, 
the  relation  between  k  and  t°  may  be  deduced  as  follows  : — 

!og  &«°  + 10  =  log  2  +  log  &«>. 
log  kt0  + 10  -  log  kt0  =  log  2 

*-a-  0-0301, 


VELOCITY  OF  BEACTION— EXAMPLES        59 

and,  on  integration, 

(1)  log  k  =  0-0301*°  +  constant. 

For  t°1%  the  temperature  corresponding  to  the  values  in  the 
preceding  problem,  A?x  =  3 '19,  and,  therefore, 

(2)  log  kl  *  0-0301*°!  +  constant. 
By  subtracting  (2)  from  (1) 

log  |  =  0-0301(*°  -  f  j). 

For  f  -  t\  =  15"  C.  we  therefore  obtain 

log  k  =  log  fcj  +  15  x  0-0301  =  0-504  +  0-452  -  0'956, 
.-.  k  =  9-04. 

If  t  and  *j  are  the  times  required  for  the  reaction  to  proceed 
to  the  same  extent  at  the  temperatures  t°  and  t0^  and  k  and 
kl  are  the  velocity-constants  at  these  temperatures,  then  if  a, 
b  and,  therefore,  since  the  reaction  proceeds  to  the  same  ex- 
tent, x  are  the  same  in  the  two  cases,  it  is  evident  from  the 
equations 


_ 

k(a  -  b)       e  a(b  -  x) 
and 

.  _1  __  I       bfa  -  x) 

1      \(a  -  b)     ^ea(b-x) 
that 

*        k 


The  times  required  for  the  reaction  to  proceed  to  the  same 
extent  at  the  two  temperatures,  are,  therefore,  ceteris  paribus, 
inversely  proportional  to  the  velocity-constants. 

For    0-002    N-NaOH,    where    ^  =  23,    ^  =  3-19    and 
k  =  9-06,  we  obtain 

q.i  q 

t  =  r_±r  x  23  -•=  8'  i  minutes, 

and  for  0*005  N-KOH,  where  t*  =  7'34,  ^  =  319  and 
k  =  9-06, 

q.-i  Q 

t  ••  M    x  7-34  =  2*6  minutes. 
9' 


60    VELOCITY  OF  EEACTION— rKOBLEMS 


Problems  for  Solution 

PBOBLEM  159. — From  the  following  data  show  that  the 
decomposition  of  H^  in  aqueous  solution  is  a  monomolecular 
reaction  : — 

Time  in  minutes,     0  10  20 

n        22-8  13-8  8-25  c.c. 

n  is  the  number  of  c.c.  of  KMnO4  required  to  decompose  a 
definite  volume  of  the  H2O2  solution. 

PROBLEM  160.  —  The  decomposition  of  AsH3  into  solid 
arsenic  and  hydrogen  may  be  followed  by  measuring  the 
pressure  at  constant  volume  from  time  to  time.  In  an  ex- 
periment at  310°  the  pressures  p  in  mm.  of  Hg  were  obtained 
after  the  times  t  hours.  Show  from  these  figures  that  the 
reaction  is  of  the  first  order. 

£  0  5-5  6-5  8       hours. 

p       733-32         305-78        818-11         835-34  mm. 

PROBLEM  161. —  The  conversion  of  acetchloranilide  into 
2?-chloracetanilide  according  to  the  equation 

C6H5 .  NC1(COCH3)  =  C6H4C1 .  NH(COCH3) 

may  be  followed  by  removing  a  measured  quantity  of  the 
solution  from  time  to  time,  adding  it  to  a  solution  of  KI, 
and  titrating  the  liberated  iodine  with  standard  thiosulphate. 
The  volume  of  thiosulphate  used  is  proportional  to  the  con- 
centration of  the  acetchloranilide.  Thus  after  t  hours  from 
the  commencement  of  the  reaction  y  c.c.  of  thiosulphate  were 
required. 

*  1  4 

y        35-6        13-8 

In  what  time  is  the  conversion  half  completed  ? 
Ans.    2-195  hours. 

PROBLEM  162. — The  decomposition  of  diacetonealcohol  into 
acetone  according  to  the  equation 

CH3 .  CO  .  CH3 .  C(CH3)2 .  OH  =  2CH3 .  CO  .  CH3 

is  accompanied  by  a  considerable  increase  in  volume.  The 
reaction  is  catalytically  accelerated  by  OH'-ions,  the  velocity- 
constant  being  proportional  to  the  concentration  of  OH'-ions. 
By  allowing  the  reaction  to  take  place  in  a  dilatometer  the 
expansion  may  be  observed.  The  quantity  of  diacetone- 


e 
VELOCITY  OF  EEACTION— PEOBLEMS    61 

alcohol  present  at  any  time  is  proportional  to  the  expansion 
from  that  time  to  the  end  of  the  reaction.  The  following 
table  gives  the  dilatometer  readings  R  at  the  times  t,  obtained 
with  a  mixture  of  20  c.c.  01  N-NaOH  and  1-0526  grams 
of  diacetonealcohol.  Calculate  the  velocity-constant  for 
0-1  N-NaOH. 

t      0    10       20         30         40         50         60         oc    mins. 
E    0   60-8   97-7    119-9    133-4    141-4    146'1    153'8 

Ans.   Mean  k  =  0-05030. 

PROBLEM  163.  —  Potassium  persulphate  and  potassium 
iodide  interact  with  liberation  of  iodine.  25  c.c.  of  a  solution, 
which  was  N/30  with  respect  to  both  persulphate  and  iodide, 
were  titrated  from  time  to  time  with  N/100  Na^SaO^  From 
the  following  results  show  that  the  reaction  is  bimolecular. 
t  is  the  time  of  titration  and  x  the  number  of  c.c.  of  thio« 
sulphate  used. 

t         9  16  32  50 

x        4-52         7-80         14-19        20-05. 

PROBLEM  164. — In  the  saponification  of  ethyl  acetate  by 
NaOH  at  10°,  y  c.c.  of  0-043  N-HC1  were  required  to  neu- 
tralise 100  c.c.  of  the  reaction  mixture  t  minutes  after  the 
commencement  of  the  reaction. 

t          0  4-89        10-37        2818  oc 

y        61-95         50-59        42-40        29-35         14-92. 

Calculate  the  velocity-constant  when  the  concentrations  are 
expressed  in  gram-molecules  per  litre. 

Ans.   Mean  k  =  2-38. 

PROBLEM  165  (cf.  preceding  problem). — 1  litre  of  N/20 
ethyl  acetate  is  mixed  at  10°  with  (a)  1  litre  of  N/20  NaOH, 
(6)  1  litre  N/10  NaOH,  (c)  1  litre  N/25  NaOH.  In  what 
time  is  half  the  ester  saponified  in  each  case  ? 

Ans.  (a)  16-8  mins.,  (b)  6'81  mins.,  (c)  24'2  mins. 
See  also  problems  319,  320. 


CHAPTER  VIII 

LAW  OF  MASS-ACTION.—  EQUILIBRIUM-CONSTANT.  —  INFLU- 
ENCE OF  TEMPERATURE  ON  EQUILIBRIUM-CONSTANT.— 
AFFINITY,  CHANGE  OF  FREE  ENERGY  OR  MAXIMUM 
WORK  OF  A  REACTION.—  PARTITION  LAW.—  SOLUBILITY 
OF  GASES. 

Law  of  Mass-action  —  Equilibrium-constant 

IF  a  number  of  substances,  for  example,  the  four  substances 
A,  B,  C  and  D,  react  according  to  the  equation 

mA  +  nB  =  pC  +  qD, 

then,  at  equilibrium,  the  following  relation  exists  between  the 
concentrations  (number  of  units  of  mass  in  unit  volume)  of 
these  four  substances  (Guldberg  and  Waage), 


The  square  brackets  denote  the  concentrations  of  the  en- 
closed substances,  m,  n,  p  and  q  are  the  number  of  mole- 
cules of  A,  B,  C  and  D  which  take  part  in  the  reaction.  Kc 
is  called  the  constant  of  the  law  of  mass-action,  or  the 
equilibrium-constant  for  the  reaction.  For  a  given  reaction, 
the  value  of  Kc  depends  on  the  temperature,  being  constant 
for  a  given  temperature,  and  on  the  units  of  mass  and  volume 
used  to  express  the  concentrations. 

If  v  is  the  number  of  units  of  volume  occupied  by  the  re- 
action mixture  at  equilibrium,  and  a,  6,  c  and  d  the  number 
of  units  of  mass  of  A,  B,  C  and  D  respectively  present  in  this 
volume  at  equilibrium,  the  concentrations  of  A,  B,  C,  and  D 
are 

[A]--*,    [B]  =  -*,[C]  =  <;,[D]=4 
62 


c 

LAW  OF  MASS-ACTION  63 

and,  from  (i), 


If  in  +  n  =  p  +  q,  that  is,  if  the  total  number  of  molecules 
on  each  side  of  the  equation  is  the  same,  equation  (2)  be- 
comes 


>  [A]m  [B]n 

that  is,  for  a  given  temperature  and  unit  of  mass,  the  value 
of  the  constant  Kc  is  independent  of  the  unit  of  volume  used 
in  expressing  the  concentrations. 

In  all  other  cases,  as  can  be  readily  seen  from  (2),  the 
value  of  Kc  is  dependent  on  the  unit  of  volume.  Thus  if  Kc  is 
the  value  of  the  equilibrium-constant  for  a  given  unit  of 
volume,  then  for  a  unit  of  volume  x  times  smaller,  the  con- 

stant   is    Ke   X    a.»i  +  n-P-q|    for 

W  =  £,    [B]  =  ^  [C]  =  £,    [D]  =  A 
and  (2)  becomes 


Thus  if  Zc  is  the  value  of  the  constant  when  the  concentra- 
tions are  expressed  in  gram-molecules  per  litre,  the  value  of 
the  constant  is  Kc  x  1000m  +  n~p~q  when  the  concentrations 
are  expressed  in  gram  -molecules  per  c.c. 

The  unit  of  concentration  usually  employed,  and  the  one 
which  will  be  used  in  the  following  examples,  is  the  gram- 
molecule  per  litre. 

That  the  value  of  Kc  depends  on  the  unit  of  mass  used  in 
expressing  the  concentrations  may  be  shown  by  expressing 
the  concentrations  in  grams  per  litre  instead  of  gram-mole- 
cules per  litre.  (2)  then  becomes 


c_c\p  (dM 
v   >   \   v 

~      c 


64  LAW  OF  MASS-ACTION 

where  MA,  MB,  Mc  and  MD  are  the  molecular  weights  of 
A,  B,  C  and  D  respectively. 

In  any  particular  case  the  unit  of  concentration  employed 
must  be  that  used  in  obtaining  the  value  of  K0  involved. 

From  (3)  it  is  evident  that  for  given  amounts  of  the  re- 
acting substances  the  position  of  equilibrium,  or  the  quantities 
of  the  reacting  substances  present  at  equilibrium,  is  inde- 
pendent of  the  volume  of  the  reaction-mixture  when  the 
same  number  of  molecules  occurs  on  each  side  of  the  reaction- 
equation.  In  all  other  cases  we  see  from  (2)  that  the 
position  of  equilibrium  is  dependent  on  the  volume  of  the 
reaction-mixture. 

In  the  case  of  gases,  their  partial  pressures  at  equilibrium 
may  be  substituted  for  their  concentrations  in  equations  (i), 
(2)  and  (3).  The  numerical  value  of  the  equilibrium-constant 
will,  of  course,  depend  on  whether  we  use  partial  pressures 
or  concentrations.  The  relation  between  the  different  values 
may  be  found  as  follows.  If  we  assume  that  A,  B,  C,  and  D 
are  all  gases,  and  that  T  is  the  absolute  temperature  of  the 
equilibrium-mixture,  then  for  concentrations  in  gram-mole- 
cules per  litre  equation  (2)  holds.  If  the  partial  pressures  in 
atmospheres  of  the  four  substances  are  pA,  pR,  pc,  pD,  then 


But  according  to  (2),  page  1,  if  a,  b,  c  and  d  are  the  numbers 
of  gram-molecules  of  A,  B,  C  and  D  present  in  v  litres  at 
equilibrium, 

aRT        __  bRT        _  cRT        _  dRT 

PA  i  PH        ~^~>  PC  ~        ~ — >  PD  —  — • — > 


•    7T    —    x    'u 
*  *     P        / ~ ~DIT\\ in f }\T)rn 


(aBT\m(bIlT\*>    (a\m(b\ 

\  v' "/  V  v  )     w  \9/ 


Only  when  p  +  q  =  m  +  n  is  Kp  —  Kc. 

If  the  equilibrium-constant  used  in  a  particular  case  is  that 
obtained  by  using  concentrations  of  the  reacting  substances 
in  gram-molecules  per  litre,  then  the  active  masses  of  the  react- 
ing substances  must  always  be  expressed  in  gram-molecules 
per  litre  when  using  this  value  of  the  equilibrium-constant. 


EQUILIBEIUM-CONSTANT  AND  TEMPEEATUEE     65 

Similarly,  if  the  equilibrium-constant  used  was  obtained  by 
employing  partial  pressures  instead  of  concentrations,  the 
active  masses  of  the  reacting  substances  must  always  be 
expressed  in  partial  pressures  when  using  this  value  of  the 
equilibrium-constant. 

If  one  or  more  of  the  reacting  substances  are  present  in 
the  solid  state,  their  active  masses,  and,  therefore,  their  con- 
centrations at  equilibrium,  are  constant,  and  may  be  omitted 
from  the  equation.  If,  for  example,  A  and  C  in  the  reaction 

mA  +  nB  =  pC  +  qD 
are  solid  substances,  the  law  of  mass-action  requires 

(s)  [-5Lq  =  K 

15;  [B]n 

Equilibrium-constant  and  Temperature 

The  equilibrium-  const  ant  Kc  or  Kp  is  dependent  on  the 
temperature  according  to  the  equations  (van't  Hoff) 


(6)  <  -    ~         and  d\ogeKp  _    -  Qp 

~~    '  ~          '         ~dT 


Qv  and  Qp  are  the  heats  evolved  in  the  reaction  from  left  to 
right  at  constant  volume  and  constant  pressure  respectively  ; 
in  the  reaction 

mA  -f  nB  =  pC  +  qD, 

for  example,  in  the  transformation  of  m  gram-molecules  of  A 
and  n  gram-molecules  of  B  into  p  gram-molecules  of  C  and 
q  gram-  molecules  of  D.  Qv  and  Qp  are  assumed  to  be  inde- 
pendent of  the  temperature  (see  p.  25).  B  is  the  gas-constant 
(1-985  or,  approximately,  2  for  calories)  and  T  the  absolute 
temperature.  On  integration  between  the  absolute  tempera- 
tures T  and  Tlt  for  which  the  respective  equilibrium-constants 
are  Kc  and  Kcl  (for  concentrations)  and  Kp  and  Kpl  (for  partial 
pressures),  we  obtain  on  changing  from  natural  to  common 
logarithms 

(7)  log*  -  log*. 

(8) 


What  was  said  about  L  in  equation  (5),  p.  25,  applies  equally 
to  Q,,  and  Qp  in  the  above  equations. 
6 


66  '  EQUILIBRIUM-CONSTANT  AND  TEMPEEATUEE 

The  following  are  some  of  the  applications  of  these  equa- 
tions.   In  all  cases  the  heat-effect  is  for  the  mean  temperature 

T  +  T 

—  x  —  l.    The  application  to  ordinary  chemical  equilibria  is 

obvious  from  the  explanations  given  above.  The  equations 
apply  also,  however,  to  many  physical  equilibria.  As  we 
have  already  seen  (p.  25),  in  the  case  of  vaporisation 

(9)  logp  -  log  ft  = 

where  p  and  pl  are  the  vapour-pressures  of  the  liquid  at  T 
and  Tl  respectively,  and  L  is  the  latent  heat  of  evaporation 
per  gram-molecule  (at  constant  pressure).  For  the  sublima- 
tion of  a  solid  the  same  equation  holds,  L  being  the  molecular 
heat  of  sublimation. 

For  the  solubility  of  solids  we  have 

Q 
(10)  logc  - 


where  c  and  cx  are  the  concentrations  of  the  saturated  solu- 
tions at  T  and  T:  respectively,  and  Q  is  the  heat  of  solution 
per  gram-molecule. 

In  the  case  of  a  difficul%  soluble  strong  binary  electrolyte, 
where  the  dissociation  in  the  saturated  solution  may  be  re- 
garded as  practically  complete,  we  have  (p.  121) 

K  =  c2  and  K^  =  c^  (solubility-products), 

where  G  and  cx  are  the  solubilities  at  T  and  2\  respectively. 
Hence 

(11)  logo*  -  logCl»  = 

where  Q  is  the  heat  of  ionisation  per  gram-molecule.  The 
heat  of  precipitation  of  the  solid  from  its  ions  has  the  same 
value  as  Q,  but  the  opposite  sign. 

For  electrolytes  which  obey  the  dilution-law  we  have 

Q     /T,  - 

(12)  log£  -  log^  =  ^B  (-%r 

where  K  and  Xj  are  the  dissociation-constants  at  T  and  Tl 
respectively,  and  Q  the  heat  of  ionisation  per  gram-molecule. 


AFFINITY— PARTITION  LAW  67 

Affinity,  or  Change  of  Free  Energy 

At  a  definite  temperature  T,  the  affinity,  or  change  of  free 
energy,  of  a  chemical  reaction,  for  example,  of  the  reaction 

mA  +  nB  =  pC  +  qD 

is 

(•3)  *  = 
or 

(14)  A  =  RT\og.K,,  +  RTlog.  ^jj^jjf 

Kc  is  the  equilibrium-constant  for  concentrations  in  gram- 
molecules  per  litre,  and  Kp  that  for  partial  pressures  in  at- 
mospheres. A  is  the  maximum  work  which  can  be  performed 
by  the  reversible  transformation  of  m  gram-molecules  of  A 
and  n  gram-molecules  of  B  at  the  concentrations  [A]  and  [B] 
gram -molecules  per  litre,  or  at  the  partial  pressures  (A)  and 
(B)  atmospheres  respectively  into  p  gram-molecules  of  C  and 
q  gram-molecules  of  D  at  the  concentrations  [C]  and  [D] 
gram-molecules  per  litre,  or  at  the  partial  pressures  (C)  and 
(D)  atmospheres. 

Partition  Law 

When  a  soluble  substance  is  distributed  between  two  non- 
miscible  solvents,  in  each  of  which  it  has  the  same  molecular 
weight,  the  ratio  of  the  concentrations  cx  and  c.2  of  the  solute 
in  the  two  solvents  has  a  constant  value  for  a  given  tempera- 
ture, or 

(15)  ~  =  S. 

C/2 

This  is  Nernst's  partition  law.  K  is  called  the  partition- 
coefficient  of  the  dissolved  substance  for  the -two  solvents. 
If  the  solute  has  not  the  same  molecular  weight  in  the  two 
solvents,  but  has,  say,  its  normal  molecular  weight  iu  the 
first  solvent,  whilst  in  the  second  solvent  it  is  more  or  less 
associated  to  complex  molecules  according  to  the  equation 


s* 

then   the   ratio  -1  is  no  longer  constant.     For  each  definite 
kind  of  molecule  S,  S«,  S,  .  .  .  Sn  however,  there  is  a  conotant 


68  SOLUBILITY  OF  GASES 

partition-coefficient  for  the  two  solvents,  and  the  ratio  of  the 
concentrations  found  experimentally  is  influenced  by  all  these. 
If  one  particular  type  of  complex,  say  Sn,  predominates  largely 
in  the  second  solvent,  the  concentration  of  the  solute  in  this 
solvent  will  be  practically  that  of  the  Sn  molecules.  Accord- 
ing to  the  law  of  mass-action,  there  exists  between  the  simple 
and  the  complex  molecules  in  the  second  solvent  the  relation 

[S\f  =  *[SJ 

and,  according  to  the  partition  law,  between  the  simple 
molecules  in  the  first  solvent  and  those  in  the  second  solvent 
the  relation 


[8], 

where  [S]j  and  [S]2  are  the  concentrations  of  the  simple 
molecules  in  the  first  and  second  solvent  respectively,  and  K 
is  the  partition-coefficient  for  this  type  of  molecule.  From 
these  two  equations  we  obtain 


S]l     -   Vfc  x  K  =  K,. 


Since  the  SM  molecules  are  supposed  to  largely  predominate 
in  the  second  solvent,  the  total  concentration  in  this  solvent 
will  be  practically  that  of  the  Sn  molecules.  If,  then,  Cj  and 
c2  are  the  total  concentrations  of  the  solute  in  the  first  and 
second  solvents  respectively  we  obtain 

(16)  JTL  =  constant. 

VC2 

cx  and  c2  may  be  expressed  in  any  unit,  for  example,  grams 
or  gram -molecules  (of  normal  molecular  weight)  per  litre. 

Solubility  of  Gases 

The  mass  of  a  gas  dissolved  by  a  given  volume  of  a  liquid 
at  a  definite  temperature  is  proportional  to  the  pressure  of 
the  gas.  This  is  Henry's  law.  Since  the  concentrations  of 
the  gas  in  the  liquid  and  in  the  space  above  the  liquid  are 
both  proportional  to  the  pressure  of  the  gas,  Henry's  law 
may  be  stated  in  the  following  form,  which  corresponds  to 
Nernst's  partition  law.  For  a  given  temperature 


SOLUBILITY  OF  GASES  69 

the  concentration  of  the  gas  in  the 

(17)    .      Uquidt    t. J-T- = (-    -  constant  =  ,. 

the  concentration  ot  the  gas  in  the 

space  above  the  liquid 

s  is  called  the  solubility-coefficient  of  the  gas  in  the  liquid 
for  the  given  temperature. 

The  solubility-coefficient  s  of  a  gas  in  a  liquid  at  a  definite 
temperature  t°  C.  may  be  defined  as  the  volume  of  gas, 
measured  at  t°  C.  and  under  a  pressure  of  p  atmospheres, 
which  is  dissolved  by  unit  volume  of  the  liquid  when  the 
pressure  of  the  gas  on  the  liquid  at  equilibrium  is  p  atmo- 
spheres. 

That  the  solubility-coefficient  s  so  defined  is  the  same  as 
the  ratio  of  the  concentrations  of  the  gas  in  the  liquid  and  in 
the  space  above  may  be  shown  as  follows  : — 

At  0°  C.  and  under  1  atmosphere  pressure,  1  gram-molecule 
of  a  gas  occupies  22*4  litres. 

1  litre  of  the  liquid  dissolves  s  litres  of  gas  measured  at  t°  C. 
and  p  atmospheres, 

273 

p  litres  measured  at  0°  C.  and  1  atmos. 


273  +  t 
273 


gram-molecules  of 
s  in  the  liquid  is 

gram-molecules  Per  litre- 


273  +  t 
/.  the  concentration  of  the  gas  in  the  liquid  is 

8  X  273?  +  t 

1  gram-molecule  of  gas  occupies  22*4  litres  at  0°  C.  and  1  atmos. 
1  gram  -molecule  of  gas  occupies 


.*.  the  concentration  of  the  gas  in  the  space  above  the  liquid 
is 


er  litre' 
and 

concentration  of  gas  in  the  liquid 
concentration  of  gas  in  space  above  liquid 
s  x  273  x  p       22-4(273  +  t)  ^ 
(273  +  Q22-4  >      273  x  p 


70  MASS-ACTION—EXAMPLES 

The  absorption-coefficient  of  a  gas  in  a  liquid  at  t°  C.,  in  terms 
of  which  the  solubility  of  gases  is  often  expressed,  is  the 
volume  of  gas,  measured  at  0°  C.  and  1  atmosphere  pressure, 
absorbed  by  unit  volume  of  the  liquid  when  the  pressure  of 
the  gas  above  the  liquid  is  1  atmosphere. 

From  a  mixture  of  gases  the  quantity  of  each  dissolved  is 
proportional  to  its  partial  pressure  at  equilibrium. 

Law  of  Mass-action  —  Equilibrium  and  Temperature  — 
Affinity  —  Examples 

PROBLEM  166.  —  a  =  9*2  grams  of  nitrogen  peroxide  occupy 
at  £x  =  27°  C.  and  under  a  pressure  of  P  =  1  atmosphere  a 
volume  vl  —  2'95  litres,  and  at  t%  =  111°  C.  and  under  the 
same  pressure  a  volume  v2  =  6*07  litres.  Calculate  the  de- 
grees of  dissociation  ax  and  a.2,  and  the  dissociation-  constants 
IJi  and  J5T2  of  nitrogen  peroxide  at  t°1  C.  and  t\  C.,  and  also  its 
molecular  heat  of  dissociation. 

SOLUTION  166.  —  According  to  the  gas  laws  the  number  n 
of  molecules  in  the  volume  vlt  at  the  temperature  tlt  and 
under  the  pressure  P,  is 

Pvl 
=  B(273-+  y 

If  M  =  92  is  the  molecular  weight  of  N2O4  and  o^  the  fraction 
of  the  N2O4  molecules  which  are  dissociated  according  to  the 
equation  N2O4  —  2NO2,  then  a  grams  of  nitrogen  peroxide 

contain       (l  -  «i)  molecules  of  N2O4  and  —      molecules  of 


NO2.     From  this  it  follows  that 

a  Pvl 


and 

MPvl 


Similarly,  for  the  temperature  t°z  C.,  we  get 

yj»i 

a2  **   /TP/O7Q   .1-    M    ~~    A* 


Substituting  the  numerical  values  given  in  the  problem, 
we  find 


MASS-ACTION—EXAMPLES  71 

92  x  1  x  2-95 

ttl  =  9-2  x  0-082  x  300   "        =  °'2°'     . 
92  x  1  x  6-07 

"2  =  w^-o-mz^m  ~    =  °'M- 

The  dissociation-constant  K  of  the  reaction 

N  A  2  2NO2 
is 

[NO2]2 

=  [NAT 

if  the  square  brackets  denote  the  concentration  in  moles  per 
litre  of  the  enclosed  molecules. 

For  ^  C.  [N204]  =  ^-(^^  and  [NOJ  =  |^, 
therefore 

X^ttnA2 

X 


a(l  -  a,)        (1  -  a 
Similarly,  for  t£  C., 


Putting  in  the  numerical  values  we  find 

4  x  9-2  x  (Q-20)2 
KI  =  0-80  x  92  x  2-95  =  °*°°68' 

_  4  x  9-2  x  (Q-93)2 
^  "  0-07  x  92  x  6-07  ~  O'Sl6' 

The  heat  of  dissociation  Q  of  a  gram-molecule  of  N2O4  can 
be  calculated  from  the  dissociation-constant  by  van't  Hoff's 
equation 

d  loge  K  _    -  Q 
dT      "  RT*' 

Integration  between  the  temperatures  Tx  =  273  +  ^  and 
T2  =  273  +  £2  gives,  if  we  assume  that  Qxdoes  not  change  in 
this  interval,  v 

'i       Q  f  1        1 


72        EQUILLBBIUM-CONSTANT—  EXAMPLES 

Substituting  the  numerical  values  gives 

1-985  x  300  x  384  0-816 

Q=  -  -       —  34-       -  x  2-3  x  log  0-^o68"=  ~  !3<>oo  cats. 

NOTE.  —  The  equation  for  the  dissociation  of  N204  may  also 
be  written 

N2O4  =  NO2  +  NO2 

and  the  dissociation-constant 

[N02]  [NOJ 
[N.OJ     ' 

where  each  NO2  molecule  is  considered  separately. 
We  have  then  (see  above)  at  tl 


and,  similarly,  at 


TN  O  1 

[N2OJ  =   M~~^~  ' 

Honce 

aa 


__ 

K*  =    (1  -  ojjfq,  =  °'2°4' 
For  a  given  temperature  either  of  the  expressions 

K  =  (1   -  a)Mv    °r   K  =    (1   -^^ 

may  be  used.  The  expression  chosen  must,  however,  be 
retained  throughout  any  series  of  calculations  dealing  with 
the  same  reaction. 

It  is  evident  that  the  value  we  obtain  for  Q  is  not  affected 
by  the  expression  chosen  for  K,  since  the  ratio  KJK^  re- 
mains constant. 

PROBLEM  167. — If  a  =  3'6  grams  of  phosphorus  penta- 
chloride  is  heated  to  t  =  200°  C.,  it  volatilises  completely,  and 
the  vapour  occupies  a  volume  v  =  1  litre  under  a  pressure 
P  =  1  atmosphere.  At  the  same  time  it  dissociates  partially 
into  phosphorus  trichloride  and  chlorine.  Calculate  the 
degree  of  dissociation  a  and  the  dissociation-constant  K  of 
phosphorus  pentachloride  at  this  temperature.  Express  the 
concentrations  in  gram-molecules  per  litre. 


MASS-ACTION—EXAMPLES  73 

SOLUTION  167.  —  As  in  problem  166  the  number  of  mole- 
cules which  occupy  the  volume  v  at  the  temperature  T°  abs. 
and  under  the  pressure  P  atmospheres  is 

Pv 


If  M  is  the  molecular  weight  of  PC15  and  a  the  degree  to 
which  PC15  is  dissociated  according  to  the  equation 

PC15  =  PC13  +  Olj,      - 
then  the  volume  v   contains    —(1  -  a)  moles  of  PC15,  — 

moles  of  PC13  and  ^  moles  of  C12.     Accordingly 

D-. 


and 

_  MPv  _  -,  208  x  1  x  1 

~~aRT  3-60  x  0-082  x  473 

and  the  dissociation-constant  at  200°  C.  is 

afa 


[PC15]  M(l  -  a)v 

(0-49)2  x  3-60 
-  208  x  0-51x1  =  °'°0815' 


PEOBLEM  168  (of.  preceding  problem). — What  pressure  P 
is  developed  when  a  =  3 '6  grams  of  solid  phosphorus  penta- 
chloride  in  v  =  1  litre  of  chlorine  at  t'  =  18°  C.  and  under 
a  pressure  p  =  1  atmosphere,  is  heated  at  constant  volume  to 
t  =  200°  C.  ? 

SOLUTION  168. — According  to  the  law  of  mass-action  the 
dissociation  of  the  PClg  is  diminished  in  presence  of  the 
gaseous  chlorine.  In  this  case  also  the  equation 

m  [PClJOy  .  K 

(1)       [PC1J 
must  be  satisfied. 

Let  a  be  the  degree  of  dissociation  of  the  PC15  at  t°  C.  in 
presence  of  the  gaseous  chlorine  ;  then  the  number  of  mole- 
cules of  .PC15  in  the  volume  v  is 


74  MASS-ACTION—EXAMPLES 

and  the  concentration  of  the  PC15  is 
(2) 


eta 
If' 


Similarly,  the  number  of  PC13  molecules  in  volume  v  is 

< 

n  = 

and  the  concentration  of  PC13 


ffi 

The  number  of  C12  molecules  is  the  sum  of  the  two  quanti- 
ties n  and  ri.  The  n  molecules  are  derived  from  the  dissociation 
of  the  PC15  and  are  equal  to  the  number  of  PC13  molecules 

=  (^LI      The   ri  molecules   are  derived   from    the   chlorine 

M 

originally  present,  and  at  t'°  C.  =  273  -f  V  =  T'  abs.  and  under 
a  pressure  of  p  atmospheres  the  volume  v  litres  contains 


ri  =         ,  molecules  of 
We  get,  therefore, 


From  (1),  (2),  (3)  and  (4)  we  obtain 

da!  (da  p     \  f  Oa          P 

MV\MV  +7tTr  )  _.  \MV 


g(l  -  a)  I  -  a 

Mv 

In  this  equation  only  a  is  unknown,  and  can,  therefore,  be 
calculated  as  follows  : — 

,        tt(a')2          a'p 

"~Mv~  +  ~BT" 
,  (  p     t       \  Mv  _  MvK 
/a  a    ' 


Mv  f  p  \          iM~v'2  f  p  \2      MvK 

«8x_V-   -^-  ,000815^ 

"   2x3-6  Vo-082  x  291   +     woio) 


MASS-ACTION—EXAMPLES  75 

7208)'     /  1~  \a     208x0-00815 


\* 


4  x  (3-6)2  VO-082  x  291  3-6 

=  -  28-85  (0-0419  +  0-00815)  +   V2T08  +  0-471 
=   -  1-442  4-   JM51 
=  0-155. 

The  total  number  of,  molecules   of  PC15  +  PC13  +  CLj  is 
therefore, 

a(l  -  a')       2aa'       pv 
~~M~        ~M  +  BT" 
3-6  x  0-845      2  x  3-6  x  0-155  1 


~T~  C\f\O  l 


208  208  r  0-082  x  291 

=  0-0146  +  0-00536  +  0-0419  =  0-0619. 

Tfhe  total  pressure  P  developed  by  heating  a  grams  of 
PC16  in  v  litres  of  chlorine,  at  p  atmospheres  and  tm  C.,  to 
200°  C.  is,  therefore, 

NET      0-0619  x  0-082  x  473 
P  = =  —  — I —  -  =  2*40  atmospheres. 

PROBLEM  169. — Nitrogen  and  oxygen  combine  at  high 
temperatures  to  form  nitric  oxide,  according  to  the  equation 

N2  +  02  =  2  NO. 
The  equilibrium-constant  at  T  =  2675°  abs.  is 

*-  3-5x10-3. 

What  yield  of  NO  (in  percentage  by  volume)  is  obtained  at 
this  temperature  and  at  normal  pressure  (1)  from  air,  (2)  from 
a  mixture  of  40  per  cent.  O2  and  60  per  cent.  N2  by  volume, 
and  (3)  from  a  mixture  of  80  per  cent.  O2  and  20  per  cent.  N2 
by  volume  ? 

SOLUTION  169. — Let  the  yield  of  NO  in  percentage  by 
volume  be  x.  Then,  since  the  reaction  takes  place  according 
to  the  equation 

N2  +  O.2  =  2  NO, 

from  a  mixture  which  originally  contained  a  per  cent,  of 
O2  and  b  per  cent,  of  N.2  by  volume,  we  get  at  equilibrium 

x  per  cent,  of  NO.  (a  -  ^)  percent,  of  O2  and  (b  -  ^j  per 
cent,  of  N2  by  volume. 


76  MASS-ACTION—EXAMPLES 

Since  the  concentrations  are   proportional   to   these  per- 
centages by  volume,   we  get  from  the  law  of  mass-action 


3-5x10-*, 


or,  solving  the  quadratic  equation  for  x, 


K(a  +  b)         lK2(a  +  fr)'2       ±abK 

and  since  K  is  small  compared  with  4,  we  may  take,  with 
sufficient  approximation, 

x  =   jKab  -  — ^-j . 

We  get,  therefore, 

(1)  for  air,  where  a  =  20'8,  and  b  =  79'2, 

x1  =  2-4  -  0-09  =  2*3  per  cent. 

(2)  for  a  =  40  per  cent.,  and  b  =  60  per  cent. 

x2  =2*8  per  cent. 

(3)  for  a  =  80  per  cent.,  and  b  =  20  per  cent. 

PROBLEM  170  (cf.  preceding  problem). — What  must  be 
the  initial  composition  of  the  mixture  of  O2  and  N2  ^to  give 
the  maximum  yield  of  NO  ? 

SOLUTION  170. — The  yield  of  NO  is 

/xr-T      E(a  +  ^) 
x  =  jKab ^-£ — '<> 

or,  since  a  +  b  =  100  in  mixtures  which  consist  of  oxygen 
and  nitrogen  only, 


x  =      #a(100  -  a)  -  K  x  25. 

To  obtain  thex  particular  concentration  a  for  which  a;  is  a 
maximum,  we  must  differentiate  x  with  respect  to  a,  and  put 
the  differential  coefficient  =  0  : 


This  expression  becomes  =  0  when  a  =  50  per  cent. 

The  yield  is  therefore  greatest  in  a  mixture  of  equal  volumes 
of  oxygen  and  nitrogen. 


MASS-ACTION—EXAMPLES  77 

PROBLEM  171.  —  On  rapidly  heating  solid  ammonium 
iodide  to  t  =  357°  C.  a  vapour-pressure  P  =  275  mm.  is  pro- 
duced, which,  owing  to  the  almost  complete  dissociation  of 
the  ammonium  iodide,  is  practically  entirely  made  up  of  the 
partial  pressures  of  the  dissociation-products  HI  and  NH3. 
On  keeping  the  system  at  this  temperature  for  some  time, 
the  vapour-pressure  increases,  because  the  hydriodic  acid 
dissociates  according  to  the  equation 

"  2  HI  =  H2  +  I2. 

What  is  the  value  of  the  vapour-pressure  P'  at  complete 
equilibrium,  if  the  dissociation-constant  for  the  reaction 
2  HI  =  H2  +  I  is 


SOLUTION  171.  —  The  total  vapour-pressure  P  is  equal  to 
the  sum  of  the  partial  pressures  of  the  individual  molecular 
species.  These  partial  pressures  may,  for  shortness,  be  ex- 
pressed by  the  chemical  symbols  enclosed  in  round  brackets. 
On  heating  rapidly  we  get  then 

P  =  (NH4I)  +  (NH3)  +  (HI), 
and  at  complete  equilibrium 

P  =  (NHJ)  +  (NH3)  +  (HI)  +  (Ha)  +  (I2). 
Since  the  dissociation  of  NH4I  into  NH3  and  HI  is  almost 
complete,  (NH4I)  may  be  neglected  in  comparison  with  (NH3) 
and  (HI),  so  that  we  obtain 

P  =  (NH3)  +  (HI) 
and 

(NH3)  =  (HI)=-f. 

For  the  dissociation  of  NH4I  vapour  the  mass-action 
equation 


must  always  be  satisfied.    If  NH4I  is  present  as  a  solid  phase, 
then  (NH4I)  is  constant  for  a  given  temperature,  and,  therefore, 


The  dissociation  of  HI  into  H2  and  I2  takes  place  accord- 
ing to  the  equation 

2  HI  =  H2  +  I2. 


78  MASS-ACTION—EXAMPLES 

At  complete  equilibrium,  reached  after  heating  for  some  time, 
the  following  equations  must,  therefore,  be  satisfied  :  — 


(1) 

(2)  (H2)  -  (I2) 

(3)  (NH3)(HI) 


(5)  P'  =  (NH3)  +  (HI)  +  (H2)  +  (I2). 

In  these  five  equations  the  five  magnitudes  P',  (NH3), 
(HI),  (I2)  and  (H2)  are  unknown.  All  the  unknowns  have, 
therefore,  to  be  calculated. 

By  eliminating  (I2)  by  means  of  equation  (2)  we  obtain 

(6)  from(l):  (NH3)  =  (HI)  +  (H2), 

(7)  from  (4)  :     (H2)  =  (HI)  JK, 

(8)  from  (5) :  P'  =  (NH3)  +  (HI)  +  2(H2). 

By  eliminating  (H2)  by  means  of  equation  (7)  we  obtain 

(9)  from  (6)  and  (7) :  (NH3)  =  (HI)  (1  +   >JK), 
(10)  from  (8)  and  (7) :  P  =  (NH3)  +  (HI)  (1+2  */K) 

(3)  divided  by  (9)  gives 

pa  PI 

4(1  +  v/£)'  "  >/l  + 

From  (11)  and  (9)  we  obtain 
P 


By  substituting  in  equation  (10)  the  values  of  (HI)  and 
(NH3)  obtained  in  (11)  and  (12),  we  obtain 


With  P  =  275  mm.  Hg. 


P'  -  137-5  x  2  +  3  X  °'123 

id/  o  x  -     ~—        =  307  mm. 


MASS-ACTION—EXAMPLES  79 

PROBLEM  172.  —  Iron  and  water  vapour  react  according  to 
the  equation 

Fe  +  H20  =  FeO  +  H2 

until  a  certain  state  of  equilibrium  is  reached.  At  tl  = 
1025°  C.,  and  under  a  total  pressure  of  1  atmosphere,  the 
partial  pressure  of  the  hydrogen  at  equilibrium  is  pH2  =  427 
mm.  Hg,  and  the  partial  pressure  of  the  water  vapour  pH20  = 
333  mm.  At  t»  =  900°  C.  the  corresponding  partial  pressures 
are  p'H2  =  450  mm.,  and  p'u20  —  310  mm.  What  is  the 
value  of  p02,  the  dissociation-pressure  of  ferrous  oxide  pro- 
duced by  its  dissociation  into  metallic  iron  and  oxygen,  at  the 
temperature  T3  =  1000°  abs.,  if  the  percentage  dissociation 
of  pure  water  vapour  at  this  temperature,  and  under  a  pres- 
sure P  =  O'l  atmosphere,  is  646  x  10~5? 

SOLUTION  172.  —  In  the  reaction  between  iron  and  steam 
according  to  the  equation 

Fe  +  H2O  =  FeO  +  KJ, 
we  have  at  equilibrium,  according  to  the  law  of  mass-action, 


where  the  value  of  K  depends  on  the  temperature  only. 
From  the  data  given  in  the  problem  K  can  be  calculated  for 
the  temperatures  T-^  =  273  +  ^  and  T2  =  273  +  ty 

We   must  assume  that  ferrous  oxide  is  dissociated  to  a 
definite,  though  small,  extent,  according  to  the  equation 

2  FeO  =  2  Fe  +  O2. 

the  degree  of  dissociation  depending  on  the  temperature. 
For  every  temperature,  therefore,  the  partial  pressure  of  the 
free  oxygen  p02,  has  a  definite  value.  This  free  oxygen  can 
react  with  the  free  hydrogen  to  form  water  vapour  according 
to  the  equation 

02  +  2H2  =  2H20. 

For  this  reaction  we  have  at  equilibrium, 


From  (1)  and  (2)  we  obtain 

1 

Po-2 


80        EQUILIBRIUM-CONSTANT—EXAMPLES 

For  every  temperature  for  which  K  and  K  are  known  we 
can,  therefore,  calculate  pOT 

K  is  known  for  the  temperatures  T1  and  T2,  and  K  for  the 
temperature  T3  can  be  calculated  from  the  dissociation  of 
water  vapour  at  that  temperature.  In  order  to  calculate  p0.2 
for  T3,  K  for  T3  must,  therefore,  first  be  found.  This  can  be 
done  by  calculating  the  heat  of  reaction,  Q,  for  the  reaction 

Fe  +  H2O  =  FeO  +  H2 
by  means  of  van't  Hofif's  equation 


dT 

Integrating  between  the  neighbouring  temperatures  Tl  and 
T2  we  obtain 

KI       ( 

and,  therefore,  » 

RT,T2  loge  ^ 
(3)  Q  =       T   _  T      2- 

For  the  temperature  interval  T3  -  T2  we  obtain  similarly 


(4)  0 

and  from  (3)  and  (4) 


,  -  ioge  K,)  -  ioge^2  -  ioga  z,. 

-l  ~     l 

Hence 

log,  ^3  =  loge^2  -  ^/y2  ~  y^gog.^  -  log.  J5g, 

^3^2   ~~    XU 

or,  converting  to  common  logarithms,  and  substituting  the 
numerical  values, 

450      1298       173  /       427  450 

log  ^3  =  log  310  ~  1000  x  125VloS333  ~  I 


=  +  0-258. 

Therefore  Ka  =  r8i. 

From  the  fact  that  water  vapour  under  a  pressure  P  =  0-1 


MASS-ACTION— AFFINITY— EXAMPLES         81 

atmosphere  is  dissociated  to  the  extent  of  6-46  x  10  ~5  per 
cent,  at  T3,  the  equilibrium-constant  K'  for  the  dissociation 
of  water  vapour  at  T3  may  be  calculated.  The  degree  of  dis- 
sociation a  =  6'46  x  10~7,  and,  since  1  molecule  H2O  gives 
1  molecule  H2  and  -J-  molecule  O2)  we  have 

PHZ  =  aP,  p02  =  %aP  and  pH2O  =  (1  -  a)  P. 
Hence 

-  a)2         2(1  -  a)2       _J2_ 
X  ^aP  ~"        a3P          "    a3P  ' 
since  a  is  very  small,  compared  with  1.     If  P  is  expressed  in 
mm.  of  Hg  as  above,  we  obtain,  therefore, 

=  1  x  1017, 


(6-46)3  x  10~21  x  76 
and 


(1-81)2  x  1017 

PROBLEM  173  (of.  preceding  problem). — What  is  the 
affinity,  in  calories,  at  T3  =  1000°  of  iron  to  oxygen  under  a 
pressure  equal  to  its  partial  pressure  in  the  atmosphere,  and 
at  what  temperature  Tx  would  ferrous  oxide,  heated  in  contact 
with  air,  dissociate  into  metallic  iron  and  oxygen,  if  the  mole- 
cular heat  of  formation  of  ferrous  oxide  is  64600  calories,  and 
is  assumed  to  be  independent  of  the  temperature  ? 

SOLUTION  173. — The  affinity  of  iron  to  the  oxygen  of  the 
air  is  measured  by  the  work  which  can  be  gained  by  the 
reversible  union  of  one  molecule  of  gaseous  oxygen  at  a 
pressure  of  1/5  atmosphere  with  metallic  iron.  For  the 
reaction 

2  Fe  +  O2  =  2  FeO 

the  equilibrium -constant  at  T3°  is  K  =  V.Po2»  where  p02  is  the 
oxygen  dissociation-pressure  of  ferrous  oxide  at  Ty 
From  equation  (14),  therefore,  at  T3 

A  =  RT3\oge  l/p02  +  ET3\oge  1/5 


From  Problem  172,  for  Ts  =  1000°,  p02  =  3-1  x  10~18  mm. 
3>1  X°"18  =  4-1  x  10-21  atmosphere. 


82          EQUILIBRIUM-CONSTANT—  EXAMPLE 

Therefore, 

A  =  2-3  x  1-985  x  1000  (log  1/5  -  log  41  x  10'21) 
=  2-3  x  1-985  x  1000  x  19-7 
=  90000  calories. 

The  relation  between  the  dissociation-pressure  of  ferrous 
oxide  and  the  temperature  is  given  by  the  equation 

geff  _     -_Q 
T       "'   RT*' 
or,  integrated  between  Tz  and  TM 

Q 


tea    s 

10«  p.-  B  \T,  ~  T 

where  pz  and  px  are  the  dissociation-pressures  at  Ts  and  Tx 
respectively.  Q  is  the  heat  of  dissociation  of  2  molecules  of 
FeO,  i.e.  the  heat  of  the  reaction  which  involves  the  forma- 
tion of  one  molecule  of  oxygen  from  2  molecules  of  FeO,  and 
therefore,  =  -  129200  calories.  If  T,  is  the  temperature  at 
which  px  =  1/5  atmosphere,  then 

41  x  IP-2*  _  129200     /I          _!  \ 

1/5  "  2-3  x  1-98  V1000  ~  T*)1 

and 


=  0-001  -  0-000696  =  0-000304, 
.'.  Tx  =  3290°  abs. 

PROBLEM  174  (using  the  results  of  the  two  preceding  pro- 
blems). —  Iron  and  carbon  dioxide  react  according  to  the 
equation 

Fe+  CO2  =  FeO  +  CO. 

At  1000°  abs.,  and  under  a  total  pressure  of  1  atmosphere 
the  partial  pressure  of  the  carbon  dioxide  at  equilibrium  is 
Pco2  —  495  mm.  Hg,  and  that  of  the  carbon  monoxide  pco  — 
265  mm.  What  is  the  degree  of  dissociation  of  pure  carbon 
dioxide  (into  carbon  monoxide  and  oxygen)  at  this  tempera- 
ture and  under  the  pressures  P1  =  01,  P2  =  1  and  P3  =  10 
atmospheres  ? 

SOLUTION  174.  —  For  the  dissociation  of  carbon  dioxide  into 
carbon  monoxide  and  oxygen  according  to  the  equation 

2C02  =  2CO  +  02, 


MASS-ACTION—EXAMPLES  83 

the  law  of  mass-action  gives  the  equilibrium  equation 


P  <;o  x 


If  the  carbon  monoxide  and  carbon  dioxide  are  in  equi- 
librium, not  only  with  the  free  oxygen  but  also  with  metallic 
iron  and  solid  FeO  according  to  the  equation 

Fe  +  C02  =  FeO  +  CO, 
the  equilibrium  equation 

(2)  £™.  =  K. 


must  also  be  satisfied. 

Finally,  the  equilibrium-pressure  of  the  free  oxygen  is  equal 
to  the  dissociation-  pressure  of  FeO  at  the  temperature  in 
question  (of.  Problem  173).  From  (2)  and  (1)  we  obtain  for 
the  dissociation-constant  of  carbon  dioxide 


For  T  =  1000°  abs. 

p02  =  4-1  x  10"iJ1  atmosphere, 


and   K  =  ^  =  0-535, 
495 

therefore, 

K  =  4-1  x  10  ~21  x  (0-535)2  =  1-17  x  10  ~21. 

From  this  value  the  degree  of  dissociation  a  of   carbon 
dioxide  under  the  pressure  P  may  be  calculated,  since 

=  (1  -  a)P  pco  =  aP  and  Poi  =  — 

From  (1)  we  obtain 

aP 
x  — 


Since  the  value  of  K'  is  very  small,  a  must  be  small  com- 
pared with  1  for  all  but  very  small  values  of  P  ;  hence 


84  MASS-ACTION—EXAMPLES 


and,    for    P1  =  O'l  atmosphere, 


tt!    =     V20    X    1<17    X    10"'21    =    2'S6    X    I0"7» 

for  P2  =  1  atmosphere, 

a2  =    «/2  x  1-17  x  IO-21  =  1-33  x  io-7, 
for  P3  =  10  atmospheres, 

a3  =  %/200  x  117  x  10 -24  =  0*616  x  io~7. 
PROBLEM  175  (cf.  preceding  problem). — Carbon  monoxide 
and  water  vapour  react  to  form  water-gas,  that  is,  a  mixture 
of  these  two  gases  with  their  reaction  products,  hydrogen  and 
carbon  dioxide.  What  is  the  composition  of  this  mixture  at 
equilibrium  at  a  temperature  of  1000°  abs.  and  under  a  total 
pressure  (1)  of  Px  =  O'l  atmosphere,  and  (2)  of  P2  =  10 
atmospheres,  if  the  carbon  monoxide  and  water  vapour  from 
which  the  equilibrium  mixture  is  produced  were  originally 
present  in  equal  volumes  ? 

SOLUTION  175. — Water  vapour  and  carbon  monoxide  react 
according  to  the  equation 

H20  +  CO  =  H.J  +  C02. 
The  mass-action  equation  at  equilibrium  is,  therefore, 


(1)  -r  "2  •  ^c°2  =  K, 

JPn2o  •  PCO 

where  pH2  etc.  are  the  partial  pressures  of  the  various  gases  at 
equilibrium.  Since  at  the  same  time  both  water  vapour  and 
carbon  dioxide  are  dissoaiated  to  a  certain  extent,  the  former 
into  H2  and  02,  and  the  latter  into  CO  and  02,  the  equations 

/n\  pj*2  •  PO%  _  JT- 

\   '     •    /«2    x  ~        1 

P   H20 

and  (3)  ^co  '  ^°2  -  ^ 


P  co2 

must  also  be  satisfied. 


Kt  and  K2  for  1000°  abs.  are  known  from  the  preceding 
examples.     By  dividing  (2)  by  (3)  we  obtain 


or  K  =  .  S 


MASS-ACTION—EXAMPLES  85 

From  problem  172  the  value  of  K±  is  1  x  10  ~  17,  if  the 
partial  pressures  are  expressed  in  mm.  Hg,  and,  from  problem 
174,  K%=  1*17  x  10  ~ 21,  if  the  partial  pressures  are  expressed 
in  atmospheres.  To  obtain  the  value  of  Kl  when  the  partial 
pressures  are  also  expressed  in  atmospheres,  1  x  10  ~  17  must 
be  divided  by  760,  giving  Kl  =  1-24  x  10  - 20. 

For  the  numerical  value  of  K  we  therefore  obtain 


24  x  10-20 


~  ^r-f.       o.n* 

"\l-17x  10-"  =:  Vl< 

The  numerical  value  of  K  is  independent  of  the  arbitrary 
unit  of  pressure,  because  both  numerator  and  denominator  of 
equation  (1)  are  of  the  same  degree  with  respect  to  p,  or,  in 
other  words,  because  the  reaction  takes  place  without  change 
of  volume  (or  pressure). 

If  the  fraction  x  of  unit  volume  of  water  originally  present 
is  transformed  into  CO2  and  H2,  which  are  formed  in  equal 
volumes,  there  will  be  in  the  equilibrium-mixture  (1  -  x) 
vol.  H20,  x  vol.  C02,  x  vol.  H2  and,  if  the  initial  volumes  of 
GO  and  H20  were  equal,  (1  -  x)  vol.  CO  ;  hence 

x  -   J?       x 

(1  -  x?          '  T~^~x  ~ 

and   x  =  0*644. 

From  a  mixture  which  originally  contained  50  °/0  by  volume 
of  H20  and  50  °/o  by  volume  of  CO,  there  is  formed,  therefore, 
an  equilibrium-mixture  containing 

32-2  70C02>  32-2  °/0H2,  17-8  °/0  CO  and  i7-870H20  by 
volume. 

Since  K  is  independent  of  the  pressure,  the  composition  of 
the  equilibrium-mixture  is  also  independent  of  the  pressure. 

PROBLEM  176  (cf.  precediag  problem).  —  The  molecular 
heat  of  combustion  of  hydrogen  is  Ql  =  58000  cals.,  and  that 
of  carbon  monoxide  is  Q2  =  68000  cals.  What  is  the  com- 
position at  equilibrium  of  the  water-gas  formed  from  equal 
volumes  of  water  vapour  and  carbon  monoxide  (1)  at  a 
temperature  Tj  =  800°  abs.,  and  (2)  at  a  temperature  T2  = 
1200°  abs.  ? 

SOLUTION  176.—  The  reaction  H2O  +  CO  =  H2  +  C02  is 
made  up  of  the  two  subsidiary  reactions 

(1)  H20  =  H2  +  iOs> 

and  (2)  £O2  +  CO  =  CO2. 


86        EQUILIBEIUM-CONSTANT—  EXAMPLES 

The  heat  of  reaction  (1)  is  Ql  =  -  58000  cals.,  and  of  reaction 
(2)  Q2  =  68000  cals.  The  heat  effect  of  the  total  reaction 
H2O  +  CO  =  H2  +  CO2  is,  therefore, 

Q  =  Ql  +  Q2  =  +  10000  cals. 

The  equilibrium-constant  K  changes  with  temperature  accord- 
ing to  the  equation 

diog.*_  -Q 

dT         RT*  ' 

Integration  between  the  neighbouring  temperatures  T  and  Tl 
gives 


and  similarly  between  the  temperatures  T  and  T2 


From  problem  175,  the  value  of  K  for  T  =  1000°  abs.  is 
3-26.  K!  for  Tt  =  800°  abs.  and  K2  for  T2  =  1200°  abs.  may, 
therefore,  be  calculated.  From  (3)  we  obtain 


0.513      0-4343  x  10000  x  200 

1-985  x  1000  x  800 
=  0-513  +  0-552  -  1-065, 
and,  therefore,  Kx  =  11*6. 
Similarly,  from  (4), 


=  0-149, 
and  K2  =  1-41 


If  xl  is  the  fraction  of  the  water  vapour  transformed  at  T-^ 
according  to  the  equation  H20  +  CO  =  H2  +  CO2,  and  x2 
the  corresponding  fraction  at  T2,  then,  as  in  problem  175,  we 
obtain 


.-*,    •-.        =  3'40' 

xl  =  0-774, 

and  -^L__  =   V^  =  VTH  =  1-19, 
i  -  ic2 

a;2  =  0-542. 


MASS-ACTION—EXAMPLES  87 

The  composition  of  the  water-gas  mixture  is,  therefore, 

at  Tl  =  800°  abs.  : 

38-7  %C02,  38-7  %H2,  1  1  -3  70H20,  1  1  -3  70CO, 
and  at  T'2  =  1200°  abs.  : 

27-1  %  C02,  27-1  °/0  H2.  22-9  %H20,  22-9  70  CO. 

PROBLEM  177.  —  A  mixture  of  a  =  10  °/0  by  volume  of  S02 
and  6  =*  90  °/0  by  volume  of  O2  is  passed  at  the  rate  of  v  =  5 
litres  per  hour  (measured  at  atmospheric  pressure  and  20°  C.) 
through  a  tube  filled  with  a  catalytic  agent  and  heated  in  a 
furnace  to  723°  C.  In  the  tube  combination  to  SO3  takes 
place,  and  we  shall  assume  that  the  gases  remain  long  enough 
in  the  tube  for  the  equilibrium  between  the  three  gases  SO2, 
O2  and  SO3  to  be  reached,  and  that  the  gases  leave  the  tube 
by  a  capillary  so  quickly  that  the  equilibrium  is  not  disturbed. 
On  leaving  the  capillary  the  gas  passes  through  a  solution  of 
barium  chloride,  from  which  it  precipitates  in  t  =  2  hours 
A  =  5  '84  grams  of  BaSO4.  What  is  the  equilibrium-constant 
Kp  of  the  reaction  2SO2  +  O2  =  2SO3  at  723°  C.  ? 

SOLUTION  177.  —  Let  the  fraction  x  of  the  SO2  present  in  the 
initial  mixture  be  converted  into  SO3  at  equilibrium.  Then, 
since^  for  every  volume  of  S03  formed  one  volume  of  SO2  and 
half  a  volume  of  O2  disappears,  a  volumes  of  SO2  +  b  volumes 
of  oxygen  give  at  equilibrium  a(l  -  x)  volumes  of  SO2  +  ax 

volumes  of  SO3  +  b  -  ^  volumes   of  oxygen.       The   total 

2 

volume  at  equilibrium  is,  therefore,  a  +  b  -  °2L.       The   total 

2 

volume   has,    therefore,   diminished    from    a  +  b  =  100    to 
a  +  b  -  ^  =  100  -  ^.     Since  the  total  pressure  remains 

constant  at  one  atmosphere,  the  partial  pressures  of  the  three 
gases  at  equilibrium  are 


(SO,)  =  -2JL2"    (SO,) 
100  -  f 

2 
atmospheres,  and,  therefore, 


88  MASS-ACTION— EXAMPLES 


(100  -  -y 


(100  -  °j)       (100  -  ™ 
#2(lOO  -  ~ 


(1  - .?  (6  -  f ) 


a;  and,  therefore,  -ZTP  may  be  calculated  as  follows  :  Let  n 
be  the  number  of  gram-molecules  of  a  gas  which  occupy  a 
volume  of  1  litre  at  20°  C.  and  atmospheric  pressure.  Then 


^  gram-  molecules  of  S02  enter  the  furnace  in  time  t.    But 
the  fraction  x  of  the   S02  is  converted  into   S03,  therefore 

xnva*  gram-molecules  of  S03  leave  the  furnace  in  time   t. 
100 

These  are  converted  completely  into  BaS04  and  weighed  as 
such.  If  M  is  the  molecular  weight  of  BaS04,  there  is, 
therefore,  precipitated  in  time  t 


A  =  "^p  grams  BaS04, 

and  (2)  z  =  !5^j_. 
nvatM 

The  equation  PV  =  W-RIT  enables  us  to  calculate  n.  Here 
P  =  1  atmosphere,  F=  1  litre,  E  =  0*082  and  T  =  (20  +  273) 
=  293.  Therefore, 


and,  substituting  the  numerical  values  in  (2), 

100  x  5-84 

0-0416  x  5  x  10  x  2  x  233  " 

Finally,  substituting  the  numerical  values  in  (1),  we  obtain 


(1  -  0-60)2  (  90   - 


10   X   0'60\         0'16   X   87 


PARTITION-LAW—  EXAMPLE  89 

Partition  Law—  Example 

PROBLEM  178.  —  The  partition-coefficient  of  iodine  for  water 
and  carbon  disulphide  is  K  =  0-0017.  An  aqueous  solution 
of  iodine  containing  a  =  O'l  gram  of  iodine  per  100  c.c.  is 
shaken  with  carbon  disulphide.  To  what  value  does  the  con- 
centration of  the  aqueous  solution  sink  (1)  when  a  litre  of  it 
is  shaken  with  50  c.c.  of  carbon  disulphide,  (2)  when  a  litre 
of  it  is  shaken  successively  with  five  separate  quantities  of 
carbon  disulphide  of  10  c.c.  each  ? 

SOLUTION  178.  —  (1)  Let  the  concentration  of  the  aqueous 
solution,  after  shaking  with  50  c.c.  of  carbon  disulphide,  be 
x  grams  per  100  c.c.  Then  10(a  -  x)  grams  of  iodine  have 
been  extracted  by  the  carbon  disulphide.  The  concentration 
of  the  carbon  disulphide  solution  is,  therefore, 
10(q  -  *)  x  100 


_ 

50 
grams  iodine  per  100  c.c.     Hence 

x          =  K 
20(a  -  x) 
0-0034 
OST  =  °'°°329  gram  per  10°  c'c- 

(2)  Let  the  concentration  of  the  iodine  in  the  water,  after 
the  first  extraction  with  10  c.c.  of  carbon  disulphide,  be  x} 
grams  per  100  c.c.  Then 


100(a  -xj 

IQOaA" 

Xl       1  +  100J5T  ' 

If  the  concentration  of  the  aqueous  solution  after  the  second 
extraction  is  x2,  we  obtain  similarly 


-  ars) 


=     /     IQQg      2 

Vl  + 


2       1  +  100#         Vl  +  lOOtf  /  ' 
After  the  fifth  extraction  the  concentration  is,  therefore, 


*17\5 
=  0-1    r.\    =*>'47  x  io~6  gram  iodine  per  100  c.c. 


The  extraction  has,  therefore,  been  practically  complete. 


90  SOLUBILITY  OF  GASES— EXAMPLE 

Solubility  of  Gases — Example 

PROBLEM  179. — The  solubility-coefficient  of  oxygen  at  0° 
is  s  =  0-04  and  of  hyrodgen  s'  =  0-02.  What  is  the  per- 
centage composition  by  volume  of  the  dissolved  gas  when  a 
litre  of  water  at  0°  is  shaken  in  a  closed  space  with  3  times 
its  volume  of  electrolytic  gas  under  an  initial  pressure  of  1 
atmosphere  ?  What  are  the  final  partial  pressures  p  and  p' 
of  the  oxygen  and  hydrogen  in  the  gas  above  the  liquid  at 
equilibrium  ? 

SOLUTION  179. — Since  J  of  the  volume  of  the  electrolytic 
gas  is  oxygen,  the  initial  partial  pressure  of  the  oxygen  in 
the  mixture  is  J  atmosphere.  Let  p  be  the  partial  pressure 
of  the  oxygen  at  equilibrium.  The  volume  of  oxygen  ab- 
sorbed by  the  litre  of  water  is 

s  =  0-04  litre  measured  at  0°  and  p  atmos. 
=.  0-04  x  3p  litre      „  „      „     J       „ 

The  oxygen  left  would  occupy  a  volume  of  3  -  0-04  x  3p 
litres  at  0°  and  J  atmos.,  but  must  fill  the  volume  of  3  litres 
at  0°  and  p  atmos.  As  the  volume  is  inversely  proportional 
to  the  pressure,  we  obtain 

3  -  0-04  x  3p  =  p      3 

3  "J 

1  -  0-04p  =  3p, 

.'.  p  =  0*329  atmos. 

The  volume  of  oxygen  absorbed  by  the  water  is,  therefore, 

=  0-04  litre  at  0°  and  0-329  atmos. 

=  0-04  x  0-329  litre  at  0°  and  1  atmos. 

=  0-0132  „          „     „    1      „ 

Similarly  for  hydrogen,  the  initial  partial  pressure  is  2/3 
atmos.,  and  the  final  partial  pressure  p'  is  obtained  from  the 
equation 

0-02  x  3p' 

3  ~  2  £_  _3p' 

3  2/3  ~=    2 

.  •.  p'  =  0*662  atmos. 
The  volume  of  hydrogen  absorbed  is 

0-02  litre  at  0°  and  0*662  atmos. 
=  0-02  x  0-662  =  0-0132  litre  at  0°  and  1  atmos, 


MASS-ACTION,  ETC.— PROBLEMS  91 

The  total  volume  of  gas  dissolved  is,  therefore,  0-0132  + 
0-0132  =  0-0264  litre  at  0°  and  1  atmos.,  and  since  it  con- 
sists of  equal  volumes  of  hydrogen  and  oxygen,  the  per- 
centage composition  is 

O2  ==  5°  Per  cent>»  H2  =  50  per  cent. 

Problems  for  Solution 
Mass-action — Equilibrium  and  Temperature— Affinity 

PROBLEM  180. — When  2-94  moles  of  iodine  and  8*10 
moles  of  hydrogen  are  heated  at  constant  volume  at.  444° 
till  equilibrium  is  established,  5-64  moles  of  hydriodic  acid 
are  formed.  If  we  start  with  5 '30  moles  of  iodine  and  7*94 
moles  of  hydrogen,  how  much  hydriodic  acid  is  present  at 
equilibrium  at  the  same  temperature  ? 

Ans.  9-49  moles. 

PROBLEM  181  (cf.  preceding  problem). — What  proportion 
of  hydriodic  acid  is  decomposed  when  1  mole  is  heated  to 
444°  till  equilibrium  is  established  ? 

Ans.  0-2197  mole. 

PROBLEM  182.— When  5-71  moles  of  iodine  and  6-22 
moles  of  hydrogen  are  heated  to  357°  till  equilibrium  is  estab- 
lished, 9-55  moles  of  hydriodic  acid  are  formed.  From  the 
equilibrium  constants  calculated  from  these  data  and  the 
data  in  problem  180,  calculate  the  heat  of  formation  of  hy- 
driodic acid  from  hydrogen  and  iodine  vapour  according  to 
the  equation 

H2  +  I2  =  2HI. 
Ans.  3028  cals. 

PROBLEM  183. — If  1  mole  of  acetic  acid  and  1 .  mole  of 
ethyl  alcohol  are  mixed,  the  reaction 

CH3 .  COOH  +  C2H6OH  =  CH3 .  COOC2H5  +  H2O 

proceeds  till  equilibrium  is  reached,  when  1/3  mole  acetic 
acid,  1/3  mole  ethyl  alcohol,  2/3  mole  ethyl  acetate,  and  2/3 
mole  water  are  present.  If  we  start  (a)  with  1  mole  acid 
+  2  moles  alcohol,  (b)  with  1  mole  acid,  1  mole  alcohol,  and 
1  mole  water,  (c)  with  1  mole  ester  +  3  moles  water,  how 
much  ester  is  present  in  each  case  at  equilibrium? 

Ans.  (a)  0-845  mole,  (b)  0'543  mole,  (c)  0-465  mole. 


92  MASS-ACTION,  ETC.— PKOBLEMS 

PROBLEM  184. — Above  150°  NO2  begins  to  dissociate  ac- 
cording to  the  equation 

N02  =  NO  +  iOa. 

At  390°  the  vapour-density  of  N02  is  19'57  (H  =  1),  and  at 
490°  it  is  18-04  Calculate  the  degree  of  dissociation  accord- 
ing to  the  above  equation  at  each  of  these  temperatures,  the 

equilibrium-constants   K  =  - — rJrk  I    ,  expressing  the   con- 

LiNU2J 

centrations  in  gram-molecules  per  litre,  and  the  heat  of  dis- 
sociation of  NO2. 

Ans.  ax  =  0-35,  a2  =  0-55.     ^  =  2-818  x  10  ~2, 
KZ  =  7-716  x  lO-2.     Q  =  -  9376  cals. 

PROBLEM  185. — At  49-7°  C.,  and  under  a  total  pressure  of 
261-4  mm.  of  mercury,  N2O4  is  63  per  cent,  dissociated  into 
NO2.     What  would  be  its  degree  of  dissociation  at  the  same 
temperature,  but  under  a  pressure  of  93-8  mm.  ? 
Ans.  80*4  per  cent. 

PROBLEM  186. — What  is  the  equilibrium-constant  at  49'7° 
for  the  above  dissociation,  (a)  for  partial  pressures  in  mm., 
(b)  for  partial  pressures  in  atmospheres,  (c)  for  concentra- 
tions in  gram-molecules  per  litre,  (d)  for  concentrations  in 
grams  per  litre  ? 

Ans.  (a)  172,  (b)  172/760,  (c)  0-00855,  (d)  01966, 

or  4  times  these  numbers,  according  to  the  expression  used 
for  the  equilibrium-constant.  See  problem  166. 

PROBLEM  187. — The  vapour-pressure  of  solid  NH4HS  at 
25*1°  is  50*1  cms.  Assuming  that  the  vapour  is  practically 
completely  dissociated  into  NH3  and  H2S,  calculate  the  total 
pressure  at  equilibrium  when  solid  NH4HS  is  allowed  to  dis- 
sociate at  25*1°  in  a  vessel  containing  NH3  at  a  pressure  of 
32  cms. 

Ans.  59'5  cms. 

PROBLEM  188. — What  is  the  total  pressure  in  the  pre- 
ceding problem  when  the  vessel  contains  H2S  at  a  pressure 
of  32  cms.  instead  of  NH3? 

Ans.  59'5  cms. 
PROBLEM  189. — In  the  reaction 

3Fe  +  4H2O  =  Fe3O4  +  4H2 


MASS-ACTION,  ETC.— PKOBLEMS  93 

there  is  equilibrium   at   200°  when  the  partial  pressure  of 

steam  is  4-6  cms.,  and  that  of  H2  is  95-9  cms.     What  is  the 

pressure  of  H2  at  equilibrium  when  that  of  steam  is  9*7  cms.  ? 

Ans.  202-2  cms. 

PROBLEM  190  (cf.  preceding  problem). — Iron  is  heated  at 
200°  in  a  closed  vessel  with  steam  at  an  initial  pressure  of 
1  atmosphere  till  equilibrium  is  established.  What  are  the 
partial  pressures  of  steam  and  hydrogen  at  equilibrium  ? 

Ans.  H2O  =  0-0458  atmos.,  H2  =  0-954  atmos. 
PROBLEM  191  (cf.  preceding  problem). — If  the  capacity  of 
the  vessel  in  the  preceding  problem  is  2  litres,  what  weight 
of  Fe3O4  is  formed  at  equilibrium  ? 

Ans.  2-85  grams. 

PROBLEM  192. — Amylene  and  trichloracetic  acid  react  to 
form  an  ester  according  to  the  equation 

001,'.  COOH  +  C5H10  =  CC13 .  COOC5Hn. 
In  an  experiment  at  100°  the  equilibrium  mixture  contained 
3*846  gram-molecules  amylene  per  litre,  0'6594  gram-molecule 
acid  per  litre,  and  2-111  gram-molecules  ester  per  litre.  If 
we  start  with  1  gram-molecule  acid,  and  4-48  gram-molecules 
amylene  in  638  c.c.  at  100°,  what  is  the  composition  of  the 
mixture  at  equilibrium  ? 

Ans.  0'174  gram-molecule  acid,  3-654  gram -molecules  amy- 
lene, and  0'826  gram-molecule  ester  in  638  c.c. 

PROBLEM  193. — Solid  NH4CN  has  a  considerable  vapour- 
pressure  at  ordinary  temperatures,  and  the  vapour  is  practi- 
cally completely  dissociated  into  NH3  and  HCN.  At  11°  C. 
the  total  vapour-pressure  is  22'7  cms.  of  mercury,  (a)  What 
will  be  the  partial  pressure  of  HCN  if  solid  NH4CN  is  al- 
lowed to  sublime  at  11°  in  a  closed  vessel  filled  with  NH3  at 
a  pressure  of  32-28  cms.  of  mercury?  (b)  What  will  be  the 
final  total  pressure  ? 

Ans.  (a)  3-595  cms.  Hg,  (b)  39'47  cms.  Hg. 

PROBLEM  194. — If  the  volume  of  the  vessel  in  the  preced- 
ing problem  is  1  litre,  how  much  NH4CN  will  sublime  ? 
Ans.  0-00203  gram- molecule. 

PROBLEM  195. — At  96°  the  ammonia  dissociation-pressure 
of  the  compound  LiCl .  NH3  according  to  the  equation 
LiCl .  NH3  =  LiCl  +  NH3 


94        EQUILIBBIUM-CONSTANT— PROBLEMS 

is  367  mm.  of  mercury,  and  at  109-2°  it  is  646  mm.      Cal- 
culate the  heat  of  dissociation  of  the  compound. 

Ans.    -  12000  cals. 

PBOBLEM  196. — At  2000°  C.,  and  under  atmospheric  pres- 
sure, carbon  dioxide  is  1*80  per  cent,  dissociated  according  to 
the  equation 

2C02  =  200  +  02. 

Calculate   the  equilibrium-constant  for   the   above   reaction 
using  partial  pressures  (in  atmospheres). 

Ans.  3  x  10  ~6. 

PBOBLEM  197. — What  is  the  equilibrium -constant  in  the 
preceding  problem,  if  the  concentrations  are  expressed  in 
gram-molecules  per  litre? 

Ans.  1-61  x  10-8. 

PBOBLEM  198. — At  28*85°  the  vapour-pressure  of  the  system 
BaCL2,  2H2O  -  BaCl2,  H2O  is  7'125  mm.,  and  at  31*65°  it  is 
8-945  mm.  Calculate  the  heat  of  hydration  of  BaCl2,  H2O 
to  BaCl2,  2H2O  by  water  vapour. 

Ans.  14910  cals. 

PEOBLEM  199. — At  30*20°  the  vapour-pressure  of  the 
system  CuS04,  5H2O  -  CuS04,  3H2O  is  10*90  mm.,  and  at 
26-30°  it  is  8-074  mm.  Calculate  the  heat  of  hydration  of 
CuSO4,  3H2O  to  CuSO4,  5H2O  by  water  vapour  per  gram- 
molecule  water. 

Ans.  13940  cals. 

PBOBLEM  200  (cf.  preceding  problem). — At  30-20°  and 
26-30°  the  vapour-pressure  of  water  is  31*93  mm.  and  25*43 
mm.  respectively.  Calculate  the  heat  of  hydration  of 
CuS04,  3H2O  to  CuSO4,  5H2O  by  liquid  water  per  gram- 
molecule  water. 

Ans.  3382  cals. 

PBOBLEM  201. — The  solubility  of  boric  acid  in  water  is 
38-45  grams  per  litre  at  13°,  and  49-09  grams  per  litre  at  20°. 
Calculate  the  heat  of  solution  of  boric  acid  per  gram- molecule. 

Ans.    -  5840  cals. 

PBOBLEM  202. — The  solubility  of  succinic  acid  at  0°  is  2-88 
gram-molecules  per  litre  and  at  8'5°  it  is  4-22  gram-molecules 
per  litre.  Calculate  the  heat  of  solution  of  succinic  acid  per 
gram-molecule. 

Ans.    -  6900  cals. 


TEMPERATURE  AND  EQUILIBRIUM— PEOBLEMS  95 

PROBLEM  203.— The  dissociation-pressure  of  CaCO3  at  810° 
C.  is  678  mm.  and  at  865°  C.  it  is  1333  mm.  Calculate  the 
heat  of  dissociation  of  CaCO3. 

Ans.    -  30110  cals. 

PROBLEM  204  (cf .  preceding  problem). — At  what  tempera- 
ture does  the  dissociation-pressure  of  CaCO3  become  equal  to 
1  atmosphere. 

Ans.   818°  C. 

PROBLEM  205  (cf.  preceding  problem).  —  What  is  the 
affinity  of  CaO  to  C02  at  atmospheric  pressure  at  a  tempera- 
ture of  810°  C.  ? 

Ans.    246  cals. 

PROBLEM  206. — At  10°  the  solubility  of  HgCl2  is  6-57  grams 
per  100  c.c.  and  at  50°  11-84  grams  per  100  c.c.      Calculate 
the  heat  of  solution  per  gram-molecule. 
Ans.     -  2669  cals. 

PROBLEM  207. — At  10°  the  electrolytic  dissociation-con- 
stants of  acetic  and  butyric  acids  are  1'79  x  10  ~ 5  and  1'66 
x  10  ~  5  respectively,  and  at  40°  they  are  1'87  x  10  ~  5  and 
1-62  x  10  ~5.  Calculate  the  heat  of  ionisation  at  25°  (a)  of 
acetic,  (b)  of  butyric  acid. 

Ans.   (a)   -  256-6  cals.,  (b)  144  cals. 

PROBLEM   208.  —  At   10°   the  ionic   product  of  water  is 
0-314  x  10  ~  14  and  at  34°  2-16  x  10  -  u.     Calculate  the  heat 
of  formation  of  H2O  from  H  and  OH'. 
Ans.   13950  cals. 

PROBLEM  209. — At  25°  the  degree  of  dissociation  of  o-chlor- 

benzoic  acid  at  a  dilution  of  512  litres  is  0'557  as  measured 

by  the  conductivity ;  at  40°  and  at  the  same  dilution  it  is 

0-521.     Calculate  the  heat  of  dissociation  of  the  acid  at  32-5°. 

Ans.   2614  cals. 

PROBLEM  210.— At  20°  the  solubility  of  AgBr  is  4-5  x  10  - * 
gram-molecule  per  litre  and  at  25°  it  is  7 '3  x  10  ~ 7.  What 
is  the  heat  of  precipitation  of  AgBr  ? 

Ans.    16880  cals. 

PROBLEM  211.— At  20°  the  solubility  of  AgCl  is  !•!  x  10  " 5 
gram-molecule  per  litre  and  the  heat  of  precipitation  of  AgCl 
is  16000  cals.     What  is  the  solubility  of  AgCl  at  30°  ? 
Ans.   1-73  x  10  ~  5  gram-mol./litre 


.96  PARTITION  LAW— PROBLEMS 

PROBLEM  212. — At  670°  C.  the  oxygen  dissociation-pressure 
of  Ba02  is  80  mm.  and  at  720°  C.  210  mm.  (a)  What  is  the 
heat  of  the  reaction  2BaO2  =  2BaO  +  O2?  (b)  At  what 
temperature  is  the  dissociation-pressure  equal  to  the  partial 
pressure  of  oxygen  in  the  atmosphere  (152  mm.)  ? 
Ans.  (a)  -  36090  cals.,  (b)  703°  C. 

PEOBLEM  213  (cf.  preceding  problem). — WThat  is  the  maxi- 
mum work  obtainable  at  670°  C.  by  the  formation  of  2  gram- 
molecules  of  Ba02  from  BaO  and  oxygen  at  the  pressure  at 
which  it  occurs  in  the  atmosphere  ? 

Ans.   1203  cals. 

PROBLEM  214. — At  2000°  C.  the  equilibrium-constant  for 
pressures  in  atmospheres  for  the  reaction  CO  -I-  £O2  =  CO2 
is  T07  x  102.  What  is  the  maximum  work  obtainable  by 
the  formation  at  2000°  C.  of  1  gram-molecule  of  -GO2  at 
atmospheric  pressure  from  1  gram-molecule  of  CO  and  -J 
gram-molecule  of  02,  both  at  atmospheric  pressure  ? 
Ans.  21210  cals. 

Partition  Law 

PROBLEM  215,.-' — At  15°  an  aqueous  solution  of  succinic  acid 
containing  0-070  gram  in  10  c.c.  is  in  equilibrium  with  an 
ethereal  solution  containing  0'013  gram  in  10  c.c.  Succinic 
acid  has  its  normal  molecular  weight  in  both  water  and  ether. 
What  is  the  concentration  of  an  ethereal  solution  which  is  in  / 
equilibrium  with  an  aqueous  solution  containing  0*024  gram 
in  10  c.c.  ? 

^-. ^Ans.   0-0044  gram  in  10  c.c. 

PROBLE^  216}— At  25°  a  solution  of  iodine  in  water  con- 
taining 0'05l#"gram  per  litre  is  in  equilibrium  with  a  CC14 
solution  containing  4-412  grams  iodine  per  litre.      The  solu- 
bility of  iodine  in  water  at  25°  is  0-340  gram  per  litre.     What  . 
is  the  solubility  in  CC14  ? 

Ans.    29 '07  grams  per  litre. 

PROBLEM  217. — In  the  partition  of  acetic  acid  between 
CC14  and  water,  the  concentration  of  the  acetic  acid  in  the 
CC14  layer  was  C  gram-mols.  per  litre  and  in  the  correspond- 
ing water  layer  W  gram-mols.  per  litre. 

C          0-292        0-363        0-725        1-07          1-41 
W         4-87          5-42          7-98          9-69        10-7 


PAETITION  LAW— PEOBLEMS  97 

Acetic  acid  has  its  normal  molecular  weight  in  aqueous  solu- 
tion. From  these  figures  show  that,  at  these  concentrations, 
the  acetic  acid  in  the  CC14  solution  exists  as  double  molecules. 
PROBLEM  218  — In  the  partition  of  succinic  acid  between 
water  and  ether  the  concentrations  of  the  acid  in  the  water 
and  ether  layers  were  c1  and  c2  respectively. 

c,         0-121         0-070         0-024     gram  in  10  c.c. 
c2        0-022        0-013        0-0046 

In  the  distribution  of  the  same  substance  between  water  and 
benzene  the  concentrations  of  the  acid  in  the  water  and 
benzene  layers  were  c3  and  c4  respectively. 

c3         0-0150       0-0195       0-0289    gram  in  10  c.c. 
c4        0-242        0-412        0-970 

Succinic  acid  has  its   normal   molecular   weight   in   water. 
What  is  its  molecular  weight  (a)  in  ether,  (b)  in  benzene  ? 
Ans.    (a)  118,  (b)  236. 

PROBLEM  219. — Phenol  has  its  normal  molecular  weight  in 
both  water  and  amyl  alcohol.  At  25°  an  amyl  alcohol  solu- 
tion containing  10'53  grams  phenol  per  litre  is  in  equilibrium 
with  an  aqueous  solution  containing  0*658  gram  per  litre. 
What  weight  of  phenol  is  extracted  from  500  c.c.  of  an  aque- 
ous solution  containing  0*4  gram-mols.  phenol  per  litre  by 
shaking  it  twice  with  amyl  alcohol,  using  100  c.c.  each  time  ? 
Ans.  17'7  grams. 

PROBLEM  220.  —  The  partition-coefficient  of  iodine  for 
CS2/water  is  410.  A  solution  of  KI  containing  8  grams  per 
litre  was  shaken  with  iodine  and  CS2  till  equilibrium  was 
established.  The  concentration  of  the  iodine  in  the  two* 
layers  was  then  determined  by  titration  with  Na2S2O3.  The 
aqueous  layer  contained  2-15  grams  iodine  per  litre  and  the 
CS2  layer  35'42  grams  per  litre.  ^Assuming  that  in  the  aque- 
ous solution  KI  reacts  with  iodine  according  to  the  equation 
KI  +  I2  =  KI3,  calculate  the  dissociation-constant  of  the  tri- 
iodide,  K  =  [KI]  [I2]/[KI3],  expressing  the  concentrations  in 
gram-mols.  per  litre.  (The  concentration  of  the  I2  in  the 
aqueous  layer  obtained  by  titration  is  the  sum  of  the  free  I2 
and  the  I2  combined  with  KI  in  the  tri-iodide,  KI  and  KI3 
are  assumed  insoluble  in  CS2.) 

Ans.  K=  1-68  x  10  ~3. 

PROBLEM  221. — In  aqueous  solution  benzoic   acid  exists- 
7 


98  PAETITION  LAW— PEOBLEMS 

as  single  molecules,  partly  electrolytically  dissociated.  In 
benzene  solution  it  exists  partly  as  single  and  partly  as 
double  molecules,  the  proportions  depending  on  the  concen- 
tration. Taking  the  partition-coefficient  water/benzene  for 
the  single  molecules  =  0-700,  calculate  from  the  following 
data  the  equilibrium-constant  for  the  dissociation  of  the 
double  molecules  into  single  molecules  in  the  benzene  solu- 
tion according  to  the  equation 

(C6H5COOH)2  =  2C6H5.  COOH. 

At  10°  the  concentration  of  benzoic  acid  in  the  water  layer 
=  0-0429  gram  per  200  c.c.  and  the  degree  of  electrolytic  dis- 
sociation =  0-169.  In  the  benzene  layer  the  concentration 
of  benzoic  acid  =  0-1449  gram  per  200  c.c.  Express  con- 
centrations in  grams  per  litre. 

Ans.  K  =  0-138. 

Solubility  of  Gases 

PBOBLEM  222. — The  solubility-coefficient  of  oxygen  in 
water  at  0°  is  0-04,  and  of  nitrogen  0'02.  If  the  composition 
of  air  by  volume  is  assumed  to  be  21  per  cent./ oxygen  and 
79  per  cent,  nitrogen,  what  is  the  percenta^e^p^mposition  by 
volume  of  the  gas  expelled  by  boiling  from  water  which  has 
been  saturated  by  free  exposure  to  air  at  0°  ? 

Ans.  34-7  per  cent.  O2.  65'3  per  cent.  N2. 

PROBLEM  223  (cf.  preceding  problem). — If  the  composition 

of  the  air  were  O2  =  20-6  per  cent.,  N2  =  79  per  cent.,  CO2  = 

0-4  per  cent,  by  volume,  and  the  solubility-coefficient  of  CO2 

*.      1-79  at  0°,  what  would  be  the  percentage/composition  of 

the  dissolved  gas  ? 

Ans.  O2  =  26-40  per  cent.,  N2  =  50-64  per  cent., 
CO2  =  22-95  per  cent. 

PROBLEM  224. — 1  litre  of  oxygen-free  water  is  shaken  in 
a  closed  space  at  0°  with  1  litre  of  oxygen  at  an  initial  pres- 
sure of  1  atmosphere,  (a)  What  volume  of  oxygen,  measured 
at  0°  and  760  mm.,  dissolves,  and  (b)  what  is  the  pressure  of 
the  oxygen  over  the  water  when  equilibrium  is  established  ? 
Ans.  (a)  38-5  c.c.,  (b)  0-9615  atmos. 

PROBLEM  225  (cf.  preceding  problem). — 1  litre  of  oxygen 
is  shaken  with  10  litres  of  water  at  0°  in  a  closed  vessel. 


SOLUBILITY  OF  GASES— PEOBLEMS  99 

The  initial  pressure  of  the  oxygen  is  1  atmosphere,     (a)  What 
volume  of  oxygen  (at  (T  and  760  mm.)  dissolves,  and  (b)  what 
is  the  final  pressure  of  the  undissolved  oxygen  ? 
Ans.  (a)  285-6  c.c.,  (b)  0'714  atmos. 

PROBLEM  226. — What  is  the  percentage  composition  by 
volume  of  the  gas  dissolved  when  a  gas  mixture  containing 
21  per  cent.  O2  and  79  per  cent.  N2  by  volume,  and  under  an 
initial  total  pressure  of  1  atmosphere,  is  shaken  in  a  closed 
vessel  at  0°  (a)  with  an  equal  volume  of  water,  (b)  with  10 
times  its  volume  of  water? 

Ans.  (a)  34-18  per  cent.  O2,  65-83  per  cent.  N2. 
(b)  31-42  per  cent.  O2,  68*58  per  cent.  N2. 

PROBLEM  227. — The  solubility-coefficient  of  CO2  at  0°  is 
1-8.     What  weight  of  CO2  under  a  pressure  of  4  atmospheres 
will  dissolve  in  a  litre  of  water  at  0°  ? 
Ans.  14-1  grams. 

PROBLEM  228. — What  is  the  relation  between  the  solu- 
bility-coefficient s  of  a  gas  at  tw  and  the  absorption- coefficient 
a  of  the  gas  ? 

a(273  +  t) 
Ans.  s  -  -V  '. 


V-* 

>ft/ 


CHAPTER  IX 

OHM'S  LAW.—  HEATING  EFFECT  OF  CUBEENT.—  FARADAY'S 
LAWS.—  SPECIFIC,  EQUIVALENT,  AND  MOLECULAR  CON- 
DUCTIVITY OF  ELECTROLYTES.—  DEGREE  OF  DISSOCIA- 
TION.— DISSOCIATION-CONSTANT.—  TRANSPORT  NUMBERS. 
—SOLUBILITY-PRODUCT.—  HYDROLYSIS 

Ohm's  Law 

A  CCORDING  to  Ohm's  law,  the  current  C  produced  in  a 
**•  conductor  of  resistance  R  by  the  potential  difference  E 
between  the  ends  of  the  conductor  is 


If  E  is  measured  in  volts,  and  R  in  ohms,  C  is  obtained  in 
amperes. 

Quantity  of  Electricity 

If  a  current  of  C  amperes  flows  through  a  conductor,  the 
quantity  of  electricity  which  passes  any  cross-section  of  the 
conductor  hi  t  seconds  is 

(2)  W  =  Ct  coulombs. 

Heating  Effect  of  Current 

The  current  C  amperes  flowing  through  the  resistance  E 
ohms  for  t  seconds,  under  the  potential  difference  E  volts,  de- 
velops the  quantity  of  energy 

(3)  Q  =  WE  =  CEt  =  C2Rt  volt-coulombs  or  joules 

=  0-239  x  C2Rt  calories 
since  1  joule  =  0*239  calorie. 

Faraday's  Laws 

Faraday's  laws  state  that  the  quantity  of  an  electrolyte  de- 
composed by  an  electric  current  is  proportional  to  the  quantity 

100 


HEATING  EFFECT;  OF-  OTEPNTEMPLES    101 


of  electricity  which  passes  any  cross-section  of  the  electro- 
lyte, and  that  the  quantities  of  different  electrolytes  decom- 
posed by  the  same  quantity  of  electricity  are  proportional  to 
their  chemical  equivalents.  These  statements  may  be 
summed  up  as  follows.  The  quantity  of  electricity,  96540 
coulombs,  is  required  for  the  decomposition  of  1  gram-equi- 
valent of  any  electrolyte,  or  for  the  liberation  of  1  gram- 
equivalent  of  any  cation  or  anion.  The  quantity  of  electricity, 
96540  coulombs,  is  called  one  faraday,  and  is  usually  denoted 
by  the  symbol  F. 

Heating  Effect—  Examples 

PEOBLEM  229.  —  A  thermostat  of  v  =  850  litres  capacity  is 
kept  at  a  constant  temperature  of  25°  C.  by  the  heat  developed 
by  maintaining  a  current  of  C  =  3'1  amp.  through  a  resist- 
ance of  E  =  22  ohms  placed  in  the  thermostat.  How  many 
degrees  does  the  temperature  of  the  thermostat  fall  in  t  =  30 
minutes,  after  switching  off  the  heating  current  ? 

SOLUTION  229.  —  If  the  thermostat  maintains  a  constant 
temperature,  the  quantity  of  heat  given  up  to  the  surround- 
ings at  any  instant  by  evaporation  of  water,  conduction  and 
radiation  of  heat,  must  be  equal  to  the  quantity  supplied  by 
the  heating  current. 

The  quantity  of  heat  supplied  by  the  heating  current  can 
be  calculated  from  the  data  given  in  the  problem.  According 
to  formula  (3),  the  heat  developed  by  the  current  in  1  second 
(t  =  1)  is  0-239  C'2B  calories.  If  the  heating  current  is 
switched  off,  the  thermostat,  therefore,  gives  up  0-239  C2E 
calories  per  second,  and  in  t  minutes  60  x  0'239  C'zRt  calories. 
If  the  thermostat  contains  v  litres  of  water  (specific  heat  ==  1), 
in  t  minutes  the  temperature,  therefore,  falls 

0_  60  x  0-23g  x  C'*Bt  _  60  x  Q-£39  x  (3-1)'2  x  22  x  30 

1000  v  1000  x  850 

=  0-107°  C. 

PROBLEM  230.  —  For  the  production  of  the  heating  current 
in  the  previous  problem  only  a  potential  of  220  volts  is  avail- 
able. A  lamp-resistance  is  to  be  used  for  the  necessary 
current-regulation.  How  many  lamps  (in  parallel)  are  re- 
quired (1)  if  each  lamp  has  a  resistance  of  JS,  =  1000  ohms  ? 
(2)  if  each  lamp  has  a  resistance  of  J52  =  300  ohms  ? 


102  FABADAY'3  LAWS* -EX  AMPLE 

E 
SOLUTION    230. — According   to  Ohm's    law   (i),    C  =  ^>. 

If,    therefore,   the    only   resistance   in  the  circuit  were  the 

220 
heating-resistance  R  =  22  ohms,  the  current  would  be  -^ty 

=  10  amps.     In  order  that  the  current  in  the  circuit  may  be 

220 
C  =  3*1   amps.,    the   resistance    must   be   increased  to-orr 

=  71  ohms.  An  additional  resistance  of  71  -  22  =  49  ohms, 
in  series  with  the  heating-resistance  is,  therefore,  required. 
This  additional  resistance  is  to  take  the  form  of  glow-lamps 
arranged  in  parallel. 

(1)  In  the  first  case  the  resistance  of  each  lamp  is  Bl  ohms. 
If  %!  such  lamps  are  arranged  in  parallel  their  total  resistance 

T* 

is  -1  ohms,  and  this  must  be  =  49  ohms, 
ni 

_RI  _  looo 

•'•  wi  -  49  ~     49  ' 

(2)  In  the  second  case  the  resistance  of  each  lamp  is  R% 
ohms.     The  total  resistance  of  w2  such  lamps  in  parallel  is 

^  ohms,  which  must  also  be  equal  to  49  ohms,  the  additional 

resistance  required, 

BZ      300 

•'•  n2  =  49  =  19 ' 

Since  nx  and  w2  must  be  whole  numbers,  the  necessary  con- 
ditions are  very  approximately  fulfilled  by 

H!  =  20  and  n2  =  6. 

In  the  first  case  the  current  is 

220 
^i  =  ctcT~t — 1000  =  ^ '06  amps. 

**  1 2TT~ 

and  in  the  second  case 

220 

^2  =  99   .    300  =  3-06  amps. 
A*  +  -*- 

Faraday's  Laws — Example 

PROBLEM    231. — A    current    passed    for    t  =  6    minutes 
through   a    voltameter   containing    dilute    HaS04    liberated 


CONDUCTIVITY— DEGREE  OF  DISSOCIATION     103 

40  c.c.  of  electrolytic  gas,  measured  at  15°  C.  and  748  mm. 
What  was  the  average  value  of  the  current  ? 

SOLUTION  231. — If  C  is  the  average  value  of  the  current  in 
amperes,  then  during  the  time  t  minutes  =  60  t  seconds 

W  =  60  Ct  =  60  x  6  x  C  coulombs  pass  through  the  solu- 
tion. At  0°  C.  and  760  mm.  the  volume  occupied  by  40  c.c. 
at  15°  C.  and  748  mm.  is 

748  x  40       273 

288       x  760  =  37'33  c'c- 

But  96540  coulombs  liberate  1  gram-equivalent  of  hydrogen 
and  1  gram-equivalent  of  oxygen.  At  0°  C.  and  760  mm. 
1  gram-equivalent  of  hydrogen  (1  gram)  occupies  ££*°o  __ 
11200  c.c.,  and  1  gram-equivalent  of  oxygen  (8  grams)  occu- 

mm.  occupied  by  the  electrolytic  gas  liberated  by  96540 
coulombs,  is,  therefore,  16800  c.c. 

Since  the  amount  of  electrolyte  decomposed  is  proportional 
to  the  quantity  of  electricity 

60  x  6  x  C  _   37-33 

96540        ~  16800' 

37-33  x  96540 

'"•  C  ~  16800  x  60  x  6  = 

Specific  and  Equivalent  Conductivity — Degree  of  Dis- 
sociation 

The  specific  resistance  of  a  conductor  is  the  resistance  in 
ohms  of  a  regular  cube  of  the  substance  of  side  1  cm.  long. 
The  reciprocal  of  this  quantity  is  the  specific  conductivity  *. 
The  unit  of  specific  conductivity  is,  therefore,  that  of  a  substance 
of  which  the  specific  resistance  is  1  ohm ;  this  unit  is  called 
the  reciprocal  ohm  or  mho,  for  which,  in  future,  we  shall 
use  the  contraction  r.o.  According  to  Ohm's  law,  the  specific 
conductivity  gives  the  current  in  amperes  produced  in  a 
regular  cube  of  1  cm.  side  when  a  potential  difference  of  1 
volt  is  applied  between  two  opposite  faces  of  the  cube. 

If  v  is  the  volume  in  litres  containing  1  gram-equivalent 
of  an  electrolyte,  the  concentration  of  which  is,  therefore, 

c  =  -  gram-equivalents  per  litre,  and  of  which  the  specific 

conductivity  is  K,  the  equivalent  conductivity  of  the  solu- 
tion is 


104  DISSOCIATION-CONSTANT 

* 

(^L.A  =  K  x  1000  v  =  —  -  —  recip.  ohms.     If  A^  is  the 

equivalent  conductivity  at  infinite  dilution,  that  is,  when  the 
dissociation  of  the  electrolyte  is  complete,  the  degree  of  dis- 
sociation in  the  solution  of  equivalent  conductivity  A  'is, 
according  to  Arrhenius, 

A 


Yelocity  of  Migration—  Ionic  Conductivity 

If  fA.A  and  fic  are  the  velocities  of  migration  of  auion  and 
cation  respectively  in  cms.  per  second  under  a  potential 
gradient  of  1  volt  per  cm.,  then  the  equivalent  conductivity 

(6)  A  =  a  x  96540  (pA  +  pc) 

(7)  -  a  ft,  +  la), 

where  1A  =  96540  ^A  and  lc  =  96540  //,e  are  the  equivalent 
ionic  conductivities  at  infinite  dilution  of  anion  and  cation 
respectively  (Kohlrausch).  When  a  —  1,  that  is,  when  the 
dissociation  is  complete,  equation  (7)  becomes 

(8)  Aw  =  1A  +  lm 

or,  in  words,  the  equivalent  conductivity  at  infinite  dilution  is 
equal  to  the  sum  of  the  equivalent  ionic  conductivities. 

Molecular  Conductivity 

The  molecular  conductivity  ^  of  a  solution  of  an  electro- 
lyte is 

i  nnn 

(9)  /A  =  1000  KV  =  -       —  recip.  ohms, 
c 

where  v  is  the  volume  in  litres  containing  1  gram-molecule 
and  c  the  concentration  in  gram-molecules  per  litre.  The 
relation  between  A  and  /x  is 

fj,  «=  ah.  and  /x,^  =  flA^, 

where  a  is  the  number  of  gram-equivalents  contained  in 
1  gram-molecule. 

Dilution  Law  —  Dissociation-constant 

If  a  is  the  degree  of  dissociation  of  a  binary  electrolyte,  and 
c  its  concentration  in  gram-equivalents  (or  gram-molecules) 


ELECTROLYTIC  DISSOCIATION—  EXAMPLES     105 

per  litre,  the  concentration  of  each  of  the  ions  is  ac  and  that 
of  the  undissociated  electrolyte  (1  -  a)c.  For  the  dissocia- 
tion of  a  binary  electrolyte  the  law  of  mass-action,  therefore, 
assumes  the  form 


(1   -  a)c  ~  (1   -  a)  -  (1   -  a)v  ~ 

v  is  the  dilution  in  litres  per  gram-equivalent  (or  gram- 
molecule)  and  .BTthe  dissociation-constant.  This  is  Ostwald's 
dilution  law.  It  holds  only  for  weak  electrolytes,  that  is, 
only  with  such  electrolytes  does  K  remain  constant  with 
varying  concentration. 

Since  a  =  -r—  ,  equation  (10)  may  be  written  in  the  form 


A2 

(")/     ""AX      -—7: —-* 

-)v         AM  (A,,,  -  A)t> 


From  these  equations  it  is  evident  that  the  numerical  value 
of  K  depends  on  the  unit  of  concentration  or  dilution  chosen. 
Here  we  shall  work  throughout  with  the  units  given  above. 

In  the  case  of  very  weak  electrolytes,  where  a  and  K  are 
very  small,  (1  -  a)  in  equation  (10)  may,  without  sensible 
error,  be  put  equal  to  1,  since  a  in  such  cases  is  very  small, 
compared  with  1.  We  thus  obtain  from  (10)  and  (n)  the 
simplified  formulas 

n 

(13)     a?C  =^  =  K, 

and 


In  a  great  many  cases  these  simplified  formulas  can  be  used 
with  sufficient  approximation. 

Electrolytic  Dissociation  —  Examples 

PROBLEM  232.  —  The  specific  conductivity  of  a  cl  =  Ol  N- 
acetic  acid  solution  at  18°  is  K^  =  0-000471*r.o.,  and  that  of  a 
c,  -  0-001  N-sodium  acetate  solution  is  K.2  =  0-0000781  r.o. 
What  is  the  dissociation  constant  K  of  acetic  acid  at  18°,  if 
the  ionic  conductivity  of  the  H'-ion  is  lc  =  318,  and  that 


106     ELECTROLYTIC  DISSOCIATION—  EXAMPLES 

of    the   Na'-ion   is-  l'c  =  44-4,    and   if  the    sodium    acetate 
solution  is  regarded  as  completely  dissociated? 

SOLUTION  232.  —  From  (4)  we   obtain   A,    the   equivalent 
conductivity  of  G1  N  -acetic  acid, 

1000  KI      1000  x  0-000471 
A.-j^--       -g^-       -=4-71r.o. 

According  to  (8),  A^,  the  equivalent  conductivity  of  acetic 
acid  at  infinite  dilution,  is  equal  to  1A  +  lc,  where  1A  is  the 
ionic  conductivity  of  the  anion  of  acetic  acid,  C2H30'2,  and 
lc  that  of  the  IT-ion.  lc  is  given,  and  1A  may  be  calculated 
from  the  equivalent  conductivity  of  the  c2  N-sodium  acetate 
solution,  which  is  equal  to  A'^,  the  equivalent  conductivity 
of  sodium  acetate  at  infinite  dilution,  as  the  salt  is  regarded 
as  completely  dissociated. 
For  sodium  acetate,  therefore, 

1000 


1000  K       1000  x  0-0000781 


.,  A 


o-ooi 

=  78-1  -  44-4  =  33-7  r.o. 
For  acetic  acid,  therefore, 

A^  =  1A  +  lc  =  33-7  +  318  =  351-7, 
and,  from  (5), 


Since  a  is  known  the  dissociation-constant  of  acetic  acid 
may  now  be  calculated  ;  according  to  Ostwald's  dilution  law 
(10) 

a2Cl        (0-0134)2  x  01 

-      L   -  '  "5 


PROBLEM  233.  —  The  velocity-constant  for  the  inversion  of 
cane  sugar  by  c  =  0*25  N-acetic  acid  at  25°  is  k  =  0-75  x  10  ~3. 
If  the  velocity-constant  is  assumed  to  be  proportional  to  the 
H'-ion  concentration,  find  the  value  of  the  constant  when 
the  acid  solution  is  also  cl  =  0*025  N  with  respect  to  sodium 
acetate,  being  giyen  that  the  dissociation-constant  of  acetic 
acid  is  K  =•=  0-000018,  and  that  the  sodium  acetate  is  dis- 
sociated to  the  extent  of  c^  =  86  per  cent. 


ELECTROLYTIC  DISSOCIATION—  EXAMPLES     107 

SOLUTION  233.  —  Let  a  be  the  degree  of  dissociation  of  the 
pure  acetic  acid  solution,  then,  according  to  (10) 

j~  ^  =  K  or   Q'25"2  =  0-000018 

(I-  a)  (1   -  a) 

.-.  a  =  0-00845. 

The  concentration  of  hydrogen  ions  in  the  solution  is,  therefore, 
[H-]  =  ac  =  0-00845  x  0-25  =  2-11  x  10  ~3  N. 

Let  a2  be  the  degree  of  dissociation  of  the  acetic  acid  in 
the  solution  containing  sodium  acetate.  The  degree  of  dis- 
sociation of  the  latter  is  ar  The  concentration  of  IT-ions 
derived  from  the  dissociation  of  the  acetic  acid  is,  therefore, 
a2c,  of  the  C2H3O'2-ions  a2c,  and  of  the  undissociated  acetic 
acid  (1  -  a2)c,  whilst  the  concentration  of  the  C2H3O'2-k>ns 
derived  from  the  sodium  acetate  is  a1c1.  The  total  concen- 
tration of  the  C2H3O'2-k>ns  is,  therefore,  a%c  +  a^.  Hence, 
if  we  use  square  brackets  to  denote  the  concentrations  of  the 
enclosed  substances, 


[C2H30'2]  _  a2c(a2C  +  qlCl) 


[C2H402] 
and,  substituting  the  numerical  values, 

a2  x  (0-25)2  x  (q2  x  0-25  +  0-86  x  0-025) 
~ 


_ 

" 


-  oj  x  0-25  l00018' 


a2  =  8-32  x  10  - 


and  [H-jj  =  a2c  =  8-32  x  10'4  x  0-25  =  2-08  x  10~4. 

If  &!  is  the  veloci 
since  the  velocity-con 
centration,  we  obtain 


If  &!  is  the  velocity-constant  for  the  mixed  solution,  then 
since  the  velocity-constant  is  proportional  to  the  H'-ion  con- 


[ill _  Q'75  x  10 -a  x  2-08  x  10  ~* 

:•]  2-11. x  io-3 

=  7-39  x  io~5. 

To  illustrate  the  degree  of  approximation  attained  by  the 
use  of  formula  (12)  instead  of  (io),  we  shall  work  the  problem 
on  the  assumption  that  a  may  be  neglected  in  comparison 
with  1. 

For  the  degree  of  dissociation  of  acetic  acid  in  the  pure 
aqueous  solution  we  obtain  from  (12) 


108     ELECTROLYTIC  DISSOCIATION—  EXAMPLES 
a2c  =  K, 

IT 


3'48xl°    ' 

and,  therefore, 

[H-]  =  ac  =  8-48  x  10-3  x  0-25  =  2-12  x  1Q-3. 

In  the  mixed  solution  we  may  assume  that  the  acetic  acid> 
in  presence  of  the  excess  of  C2H30'2-ions  from  the  dissociation 
of  the  sodium  acetate,  is  practically  undissociated,  that  is, 
[C2H4O2]  =  c,  and  that  the  C2H3O'2-ions  are  furnished  only 
by'  the  sodium  acetate.  Therefore  [C2H3O'2]  =  a1c1  and 

[O.H,0-J 


_ 


[C.HA] 
Kc       0-000018 


.  P*].  -  a^;  -     0-86  x  0-025     '  2'09  *.1(  -• 

Since, 

k-v 

0-75  x  10-3  x  2-09  x  1Q-4 
-.-  2-12  x  10 -~  =  r4  X  '°     ' 

PROBLEM  234. — The  angle  of  rotation  of  an  a  =  5  per  cent, 
cane  sugar  solution  in  a  tube  20  cms.  long  is  A0  =  +  66*7°. 
After  complete  inversion  the  angle  of  rotation  is  A^  =  -  19*7°. 
If  the  solution  contains  in  addition  cx  =  O'Oi  N-HC1,  the 
angle  of  rotation  diminishes  by  69 '2°  in  t  =  20  minutes.  By 
how  much  does  the  angle  of  rotation  diminish  in  the  same 
time,  when,  instead  of  HC1,  the  solution  contains  c2  =  O'l  N- 
lactic  acid,  if  the  dissociation- constant  of  lactic  acid  is 
K  =  1*4  x  10~4?  Assume  the  HC1  to  be  completely  dis- 
sociated. 

SOLUTION  234. — If  A0  denotes  the  initial  angle  of  rotation, 
before  the  inversion  begins,  and  Ax  the  final  angle  of  rota- 
tion after  complete  inversion,  the  total  change  in  the  angle 
of  rotation,  A0  -  AM,  is  proportional  to  the  initial  amount  of 
cane  sugar,  a  per  cent.  Similarly,  if  at  the  time  t,  when  the 
solution  contains  x  per  cent,  of  cane  sugar,  the  angle  of  rota- 
tion is  Ax,  the  change  in  the  angle  of  rotation,  Ax  -  A^,  is 
proportional  to  x.  Therefore, 


ELECTROLYTIC  DISSOCIATION—  EXAMPLES     109 

Since  in  the  presence  of  a  large  excess  of  water  the  inver- 
sion proceeds  as  a  monomolecular  reaction,  the  velocity  of 
inversion  is  given  by  the  equation 

dx 


where  k  is  the  velocity-constant  of  the  reaction,  and  x  the 
concentration  of  the  cane  sugar  at  the  time  t.  On  integrating 
we  obtain 

(2)  logex  =  -  kt  +  constant. 
When  t  =  0,  x  =  a, 

(3)  /.  loge  a  =  constant. 
Subtracting  (2)  from  (3)  we  obtain 


and,  therefore,  from  (1) 


or,  changing  from  natural  to  common  logarithms, 


The  inversion  is  catalytically  accelerated  by  the  H'-ions, 
and  the  velocity-constant  k  is  proportional  to  the  IT-ion 
concentration.  If  we  assume  the  HC1  to  be  completely  dis- 
sociated, the  concentration  of  the  H'-ions  is  [H*^  =  cr  The 
velocity-constant  k-±  for  this  particular  case  may  be  calculated 
from  the  data  given  in  the  problem  by  means  of  equation  (4) 
which  in  this  case  becomes 


In  the  c2.  N-lactic  acid  let  the  H'-ion  concentration  be 
[H*]2.  If,  for  brevity,  we  represent  lactic  acid  by  the  formula 
LH,  where  L  is  the  anion,  we  obtain  for  the  dissociation  of 
lactic  acid  the  equilibrium-equation 


110    ELECTKOLYTIC  DISSOCIATION— EXAMPLES 

EiU5i  =  ^ 

or,  since  [I/]  =  [H-]2, 
[H-V 


and  [H-],  =  -  j 

If  k2  is  the  velocity-constant  for  the  inversion  of  cane  sugar 
by  c2  N-lactic  acid,  we  obtain,  since  the  velocity-constant  is 
proportional  to  the  H'-ion  concentration, 


T  *  -  **  ~  1 


In  this  equation  only  &2  is  unknown,  and  can,  therefore,  be 
calculated. 

The  angle  of  rotation  Av  in  the  lactic  acid  solution  after 
t  minutes  is  obtained  from  the  equation  (cf.  (4)) 


Substituting  the  numerical  values  in  the  various  equations 
we  obtain  klt  &2  and  Ay,  and,  therefore,  the  diminution  of  the 
angle  of  rotation  required,  A0  —  Ay.  Thus  in  the  case  of  the 
inversion  by  HC1,  since  the  diminution  of  the  angle  of  rotation 
after  20  minutes  is  69*2°,  we  obtain  Af  =  66T  -  69'2°  = 
-  2-5°,  and  from  equation  (5) 

2-3 

*i  =  7- 

2-3.       66-7  +19-7  K1     86-4 

=  20  l0^  -  2-5  +  19-7  =  °115  lo§  1T2 
=  0-115  x  0-703  =  0-0808. 
From  (6) 
fc, 


0-1 


ELECTROLYTIC  DISSOCIATION—  EXAMPLES     111 

=  8-08(VO-49  x  10-8  +  1-4  x  10~6  -  0'7  x  10"*) 
=  8-08(3-74  x  10-3  -  0-7  x  10~4) 
=  0-0297. 
From  (7) 


0-0297  x  20 
=  log  86-4-       —  c^j— 

=  1-937  -  0-258  =  1-679, 
.-.  Ay~Aca=  47'8'J, 
and  Ay  =  47'8  +  AM  =  47'8  -  19-7  =  28-1°. 

The  diminution  of  the  angle  of  rotation  after  20  minutes  in 
the  lactic  acid  solution  is,  therefore, 

A0  -  Ay  =  66-7°  -  28-1°  =  38*6°. 

PROBLEM  235  (cf.  preceding  problem).  —  What  is  the  dis- 
sociation-constant of  acetic  acid  if  a  mixture  of  c3  =  0'05  N- 
lactic  acid,  and  c4  =  0*38  N-acetic  acid  has  the  same  inverting 
action  as  the  c2  =  O'l  N-lactic  acid  alone  ? 

SOLUTION  235.  —  If  a  solution  containing  c3  N-lactic  acid 
and  c4  N-acetic  acid  has  the  same  inverting  action  as  a  c2  N- 
lactic  acid  solution  alone,  the  H*-ion  concentrations  of  these 
two  solutions  must  be  the  same,  since  the  inverting  action  is 
proportional  to  the  H--ion  concentration. 

If,  for  brevity,  we  represent  lactic  and  acetic  acids  by  the 
formulae  LH  and  AH,  where  L  and  A  are  the  respective 
anions,  we  obtain  for  the  electrolytic  dissociation  of  lactic 
acid 


and  of  acetic  acid 


where  K  and  K'  are  the  dissociation-constants  of  lactic  acid 
and  acetic  acid  respectively.  As  in  Solution  234  we  have 
also 


(3) 


112     ELECTROLYTIC  DISSOCIATION— EXAMPLES 

and  in  the  mixed  solution 

(4)  [L']  +  [LH]  -  cs, 

(5)  [A']  +  [AH]  -  ev 
and 

(6)  [L']  +  [A']  =  [H-]. 

From  these  six  equations  the  six  unknown  quantities 

[L'],  [A'],  [LH],  [AH],  [H-]  and  K  may  be  calculated. 
From  (3) 

-^T3      8+  1-4x10-6-        -^ 

=  3-67  x  10-3, 
and  from  (4) 

[L']  =  c3  -  [LH]  =  0-05  -  [LH]. 

Substituting  these  values  of  [IT]  and  [L']  and  the  numeri- 
cal value  of  K  in  (1)  we  obtain 

(0-05  -  [LH])  x  3-67  x  10  ~3  ==  1-4  x  10  ~4  [LH], 


and,  therefore,  from  (4) 

[L']  =  0-05  -  0-04818  =  0-00182. 
From  (6) 

[A']  =  [H-]  -  [L']  =  0-00367  -  0-00182  =  0-00185. 
From  (5) 

[AH]  =  c4  -  [A']  =  0-38  -  0-00185  =  0-378, 
and,  finally,  from  (2) 


K/-     [AH] 

_  000185  x  0-00367 

0-378 
=  i '8  x  10  ~5. 

PROBLEM  236. — In  a  mineral  water  with  an  acid  reaction 
the  total  concentration  of  carbonic  acid,  free  and  -combined, 
is  a  =  0-620  gram  CO2  per  litre,  the  total  concentration  of 
sulphuretted  hydrogenffree  and  combined,  is  b  =  0*018  gram 
H2S  per  litre,  and  the  concentration  of  sodium  oxide  is  c  = 


ELECTEOLYTIC  DISSOCIATION— EXAMPLES  113 

0-279  gram  Na2O  per  litre.  What  is  the  concentration  of  free 
carbonic  acid  [H2CO3]  and  of  free  sulphuretted  hydrogen 
[H2S],  if  the  first  dissociation-constant  of  H2CO3  is  K^  = 
3-04  x  10-7,  and  that  of  H2S  is  0'91  x  10  ~  7  ? 

SOLUTION  236. — In  the  acid  mineral  water  the  carbon 
dioxide  is  present  as  undissociated  molecules  of  carbonic 
acid  H2CO3  and  as  the  anion  HCO'3,  an(^  ^ne  hydrogen  sul- 
phide is  present  as  undissociated  H2S  and  as  the  anion  HS'. 

If  M1  is  the  molecular  weight  of  C02  and  M%  that  of  H2S 
then 

(1)  [H2C03]  +  [HCO'J  =  Jl, 

(2)  [H,S]  +  [HS']  =  A, 

where  the  square  brackets  denote,  as  usual,  the  concentra- 
tions of  the  enclosed  substances. 

From  the  law  of  electro-neutrality,  according  to  which  the 
total  number  of  equivalents  of  cations  must  be  equal  to  the 
total  number  of  equivalents  of  anions  in  the  solution,  it 
follows  that 

(3)  [HC03']  +  [HS']  =  [Na-]  =  ^, 

where  M3  denotes  the  molecular  weight  of  Na2O.  For  the 
solution  of  these  three  equations  with  the  four  unknowns 
[H2CO3],  [H2S],  [HCOg']  and  [HS'],  a  fourth  equation  is 
required.  This  is  obtained  by  an  application  of  the  law  of 
mass-action. 

For  the  ionic  dissociation  of  the  two  weak  acids  the  fol- 
lowing equations  hold, 

[HC03']  [H-]  =  K,  [H2C03] 
and  [HS']  [H>]  =  K,  [H2S], 

or,  dividing  the  latter  by  the  former, 

(i) 
}  [HOOT] 

The  problem  is,  therefore,  capable  of  solution. 

From  a  consideration  of  the  numerical  values  given  in  the 
problem,  it  follows  that  [HS'J  is  small  compared  with  [HCO/J, 
especially  as  hydrogen  sulphide  is  the  weaker  acid.     Neglect- 
8 


114     ELECTROLYTIC  DISSOCIATION— EXAMPLES 

ing  [HS']  in  comparison  with  [HC03'J,  we  therefore  obtain 
from  (3) 


[HCO3']  =  j£  =  -     6o         =  O'OO9  gm.-ion  per  litre, 

x    3  Va 

and  from  (1) 

[H2CO3]  =  4-  -  0-009  =  0-0141  -  0-009  =  0-0051  mole 
i 

per  litre. 

From  (4)  we,  thus  obtain 

[HS']  _  0-91  x  10-T       0-009 
W   [H2S]~  3-04  x  10 -7  X  0-0051  , 

and  [HS']  =  0-53  [H2S].     Substituting  this  value  of  [HS']  in 
equation  (2)  we  have 

[H2S]  (1  +  0-53)  -  jf*  -  0-00053, 

/.  [H2S]  =  0*000346  gram-molecule  per  litre, 
and 

[HS']  =  0-000184  gram-ion  per  litre. 

PROBLEM  237. — When  picric  acid  distributes  itself  between 
water  and  benzene,  only  the  undissociated  molecules  of  the 
acid  dissolve  in  the  benzene,  and  the  partition-coefficient  of 
the  undissociated  molecules  for  benzene/water  is  K^  =  1/0-0281. 
The  dissociation-constant  of  picric  acid  in  water  is  K2  =  0-164. 
What  is  the  concentration  c0  of  an  aqueous  solution  of  picric 
acid  which  is  in  equilibrium  with  a  c  =  0-07  N-solution  of 
picric  acid  in  benzene? 

SOLUTION  237. — Picric  acid  is  soluble  both  in  water  and  in 
benzene ;  if  an  aqueous  solution  of  picric  acid  is  shaken  with 
benzene,  the  picric  acid  distributes  itself  between  the  two 
solvents  and  the  ratio  of  the  concentrations  of  the  undis- 
sociated picric  acid  molecules  in  the  two  solvents  has  a  con- 
stant value  Kv  which  is  independent  of  the  concentration. 
The  aqueous  solution  conducts  the  electric  current,  and, 
therefore,  contains  the  ions  of  picric  acid  besides  undissociated 
molecules ;  the  benzene  solution,  on  the  other  hand,contains 
no  ions  since  it  is  a  non-conductor. 

Let  the  total  concentration  of  an  aqueous  solution,  which 
is  in  equilibrium  with  a  c  N-benzene  solution,  be  c0  =  cl  +  c2, 
where  c1  is  the  concentration  of  the  undissociated  molecules 


ELECTEOLYTIG  DISSOCIATION—  EXAMPLES     115 

and   c2    that   of   the    ions.      At   equilibrium    the    following 
equations  must  hold  :  — 

(1)  from  the  partition  law,  —  =  Kv 

ci 

c  2 

(2)  from  Ostwald's  dilution  law  —  =  K2. 


From    these    two    equations    and    the    further  equation 

=  cl  +  c2,  the  three  unknowns 

la  ted. 

By  dividing  (1)  by  (2)  we  obtain 


c0  =  cl  +  c2,  the  three  unknowns  c0,  cl  and  c2  may  be  cal 
culated. 


fOL 

and  c2  =  \j  —•', 
Ai 


from  (1) 

therefore 
c0  = 


=  0-07  x  0-0281  +  jQWTx  0-0281  x  0-161 

=  0*0200  N. 

PROBLEM  238. — A  methyl  acetate  solution  is  saponified  by 
a  cl  =  O'l  N-solution  of  KCN  a  =  3'3  times  as  fast  as  by  a 
c2  =  0-01  N-solution.  What  is  the  dissociation-constant  K 
of  hydrocyanic  acid,  if  the  ionic  product  of  water  is  Kw  = 
0-8  x  10  ~14? 

SOLUTION  238. — The  solution  of  KCN  is  alkaline  on  ac- 
count of  hydrolysis,  and  in  alkaline  solution  the  saponification 
of  the  ester  takes  place  according  to  the  equation 

CH3COOCH3  +  OH'  «  CH3OH  +  CH3COO'. 

The  velocity  of  saponification  is,  therefore,  proportional  to 
the  product  of  the  concentrations  of  the  ester  and  of  the  free 
hydroxyl  ions.  In  two  solutions,  in  which  the  concentrations 
of  the  ester  are  equal,  the  velocities  of  saponification  are, 
therefore,  proportional  to  the  concentrations  of  the  hydroxyl 
ions  [OH']!  and  [OH']2.  In  the  present  case  we  have, 
therefore, 


116    ELECTROLYTIC  DISSOCIATION—  EXAMPLES 


[OH']2 

From  the  law  of  mass-action  we  obtain  for  the  dissociation 
of  hydrocyanic  acid 

[ONI  [H-]  _  K 
[HCN] 

and,  since  [H-]  [OH']  =  Kw, 
K 


The  concentration  of  the  undissociated  KCN  may  be  neglected, 
since  the  salt  may  be  regarded  as  completely  dissociated. 
The  concentration  of  the  K'-ions  is,  therefore,  practically 
equal  to  the  total  concentration  c  of  the  KCN.  The  principle 
of  electro-neutrality  requires 

c  =  [K-]  =  [ON']  +  [OH'], 

as  the  hydrolysis  of  KCN  takes  place  according  to  the  equation 
KCN  +  H2O  =  HCN  +  KOH, 

and  the  KOH  like  the  KCN  may  be  regarded  as  completely 
dissociated. 

We  obtain,  therefore, 

(3)  o,  =  [ON'],  +  [OH'],, 
and  (4)  c,  -  [ONI  +  [OH1],  =  [CN']2  +  [^Ji  (from  (1)). 

For  the  solution  of  the  three  equations  (2),  (3)  and  (4),  in 
which  the  five  quantities  #,  [HCN],  [CN']1}  [CN']2  and  [OH'^ 
are  unknown,  we  require  two  further  equations.  These  we 
obtain  as  follows.  The  concentration  of  the  OH'-ions  pro- 
duced by  the  hydrolysis  is  equal  to  the  concentration  of  the 
undissociated  HCN  produced  at  the  same  time  according  to 
the  equation 

ON'  +  H2O  =  OH'  +  HCN. 
Therefore,  from  (2), 

I  /TM''"]  T7" 

[HCN]  =  [OH']  -  LJ  x 
and 


ELECTKOLYTIC  DISSOCIATION—  EXAMPLES     117 


and 


By  dividing  (5)  by  (6)  we  obtain 

ESl_ 

[ONI, 
and,  from  (4), 

(7)C2 

From  (3)  and  (7)  we  obtain 


a  a*  a 

.-.[CN-],-^- 

and 

[OH'],  =  Cl  -  [ON'], 
Finally,  from  (5), 
K  = 


*±:*m 


_        o(c1  -  c2a)  (a  -  1) 

~  (a'2c0  —  c  )2 

_  0-8  x  10-"  x  3-3  x  2-3  x  (0-1  -  Q-Q33) 

{(3-3)2  x  0-01  -  O-l}2 
=  5  x  io-". 

Transport  Numbers 

If  /JLA  and  p,c  are  the  velocities  of  migration  of  the  anion 
and  cation  respectively,  the  fraction  -  $  —  =  n    is    called 

PA  +  Po 

the  transport  number  (migration  number)  of  the  anion,  and  the 

LL  '  '  * 

fraction  -  ^—  =  1  —  "  n  the  transport  number  of  the  cation 

PA   +  PC 

(Hittorf).     Of  the  total  quantity  of  electricity  which  passes 
any  cross-section  of  an  electrolyte,  the  fraction  n  is  carried 


118  TKANSPOKT  NUMBERS 

by  the  anious  or  negative  ions  and  the  fraction  1  -  n  by  the 
cations  or  positive  ions.  From  equations  (6),  (7)  and  (8)  it 
is  evident  that 

n  =  7— IT  =  A^ 

I  A  +  lc        Aoo 

so  that  from  a  knowledge  of  n  and  A^  we  can  obtain  1A  and 

lc. 

n  is  determined  experimentally  by  electrolysing  a  solution 
of  the  electrolyte  in  question,  measuring  the  total  quantity  of 
electricity  which  has  passed  through  the  solution  (e.g.  by  a 
silver  voltameter  in  series  with  the  electrolyte)  and  the 
changes  of  concentration  at  the  electrodes.  If  the  quantity 
of  electricity  F  is  passed  through  the  solution,  1  gram-equiva- 
lent of  anion  is  deposited  at  the  anode,  n  gram-equivalent  of 
anion  migrates  to  the  anode,  and  1  -  n  gram-equivalent  of 
cation  migrates  from  the  anode,  that  is,  at  the  anode  1  —  n 
gram-equivalent  of  both  anion  and  cation  disappears,  or  1  -  n 
gram-equivalent  is  the  fall  in  concentration  of  the  electrolyte 
round  the  anode.  Similarly  at  the  cathode,  1  gram-equivalent 
of  cation  is  deposited,  1  -  n  gram-equivalent  of  cation  migrates 
to  the  electrode,  and  n  gram-equivalent  of  anion  migrates 
from  the  electrode,  that  is,  at  the  cathode  n  gram-equivalent 
of  both  cation  and  anion  disappears  from  the  solution  round 
the  electrode,  or  the  fall  in  concentration  of  the  solution 
round  the  cathode  is  n  gram-equivalent.  If  then-  we  assume 
that  the  changes  in  concentration  round  the  electrodes  are 
due  only  to  discharge  and  migration  of  the  ions,  that  no 
secondary  changes,  such  as  solution  of  the  electrodes,  etc., 
take  place,  and  that  no  diffusion  occurs,  we  have  the  relation 

fall  of  cone,  of  electrolyte  round  cathode          n 
fall  of  cone,  of  electrolyte  round  anode      ~  1  -  n 

Since  the  quantities  on  the  left-hand  side  of  the  equation 
are  known,  n  may  be  calculated. 

It  is  not,  however,  necessary  to  determine  the  change  in 
concentration  at  both  electrodes.  If  the  theoretical  change 
(fall)  in  concentration  in  gram-equivalents  at  one  electrode, 
say  the  cathode,  and  the  total  quantity  of  electricity  in 
coulombs  passed  through  the  solution  (e.g.  by  the  silver 
voltameter)  are  determined,  all  the  necessary  data  are  obtained. 
Then 


TKANSPOET  NUMBERS— EXAMPLES         119 
fall  (in  gram-equivs.)  round  cathode  x  96540 

or 


total  quantity  of  electricity  passed 


fall  (in  gram-equivs.)  round  anode  x  96540 
total  quantity  of  electricity  passed 

Since,  however,  the  experimental  arrangements  for  the  deter- 
mination of  transport  numbers  vary  with  the  nature  of  the 
ions  under  investigation,  and  secondary  reactions  at  the  elec- 
trodes, which  vary  from  case  to  case,  have  to  be  taken  into 
account,  the  calculation  for  each  particular  case  has  to  be 
thought  out  independently. 

Transport  Numbers—  Example 

PEOBLEM  239.  —  A  solution  containing  0-1605  per  cent. 
NaOH  was  electrolysed  between  platinum  electrodes.  After 
electrolysis  55-25  grams  of  the  cathode  solution  contained 
0*09473  grams  NaOH,  whilst  the  concentration  of  the  middle 
portion  of  the  electrolyte  was  unchanged.  In  a  silver  volta- 
meter in  series  the  equivalent  of  0*0290  gram  NaOH  was 
deposited  during  electrolysis.  Calculate  the  transport  num- 
bers of  the  Na*  and  OH'  ions. 

SOLUTION  239.  —  After  electrolysis  55  '25  grams  of  the 
cathode  solution  contained  0*09473  gram  NaOH  or  (55-25  - 
0-09473  =  )  55*155  grams  of  water  contained  0-09473  gram 
NaOH. 

Before  electrolysis  100  grims  of  the  cathode  solution  con- 
tained 0*1605  gram  NaOH,  therefore,  55*155  grams  water 


contained  n  =  0*08867  gram  NaOH., 

' 


The  increase  in  the  amount  of  NaOH  round  the  cathode  is 
therefore,  0-09473  -  0-08867  =  0-00606  gram. 
At  the  cathode,  however,  the  secondary  reaction 

Na  +  H20  =  NaOH  +  H   ' 

takes  place,  the  discharged  Na'-ions  reacting  with  the  water 
to  re-form  their  equivalent  of  NaOH.  Had  no  secondary 
reaction  taken  place,  therefore,  the  cathode  solution  would 
have  contained  0*0290  gram  NaOH  less  'than  it  did,  since 
0*0290  gram  NaOH  is  the  equivalent  of  the  Nations  dis- 
charged, as  measured  by  the  silver  voltameter.  The  amount; 
of  NaQH  in  trie  cathode  sqlutiqn  would,  therefore,  have  been 


120         TRANSPORT  NUMBERS—  EXAMPLES 

0-09473  -  0-0290  =  0-06573  gram  in  55-155  grams  water, 
compared  with  0-08867  gram  before  electrolysis.  The  theo- 
retical fall  in  concentration  round  the  cathode,  that  is,  on  the 
assumption  that  no  secondary  reaction  had  taken  place,  is, 
therefore,  0-08867  -  0-06573  =  0-02294  gram  NaOH.  If  M 
is  the  equivalent  weight  of  NaOH  we  obtain 

_  fall  in  gram-equivs.  round  cathode  x  96540 

total  quantity  of  electricity  passed 
0-02294       qfi540 
~M-  X  9654°  _  0-02294  _ 

96540       "  0'0290   = 


M 
and  i  -  n  =  1  -  0-791  =  0*209. 

Alternative  Method.  —  For  the  deposition  of    _  gram- 

equivalent  of  silver  in  the  voltameter,  the  increase  in  the 

0*00606 
amount  of  NaOH  round  the  cathode  is  -  gram  -  equiv. 

Therefore,  for  the  passage  of  1  faraday  (96540  coulombs), 
corresponding  to  the  deposition  of  1  gram-equivatent  of  silver 
in  the  voltameter,  the  increase  in  the  amount  of  NaOH  round 
the  cathode  would  have  been 

0-00606  1         =  0-00606          M      =  Q-QQ606  gram- 

M        <  0-0290/M  M       X  0-0290       ¥0290"  equiv. 

But  for  the  passage  of  1  faraday  n  gram-  equivalent  of  OH' 
migrates  from  the  cathode,  1  -  n  gram-equivalent  of  Na- 
migrates  to  the  cathode  and  1  gram-equivalent  of  Na*  is  dis- 
charged at  the  cathode.  The  theoretical  fall  in  the  amount 
of  electrolyte  round  the  cathode  for  the  passage  of  1  faraday 
is,  therefore,  n  gram-equivalent  of  OH'  and  1  -  (1  -  n)  =  n 
gram-equivalent  of  Na-,  or  n  gram-equivalent  of  NaOH.  But 
owing  to  the  secondary  reaction  of  the  discharged  gram- 
equivalent  of  Na-  with  the  water,  1  gram-equivalent  of  NaOH 
is  re-formed.  Instead  of  a  fall  of  n  gram-equivalent  there  is, 
therefore,  an  increase  round  the  cathode  of  1  -  n  gram- 
equivalent  of  NaOH.  Therefore 


and  n  =  1  -  0-209  =  0791, 


SOLUBILITY-PRODUCT  121 

Solubility  and  Solubility-product 

From  the  law  of  mass-action  it  follows  that  for  a  saturated 
solution  of  a  difficultly  soluble  electrolyte  Am  BM  in  contact 
with  the  solid  phase 

(14)  [A-]"  [B']«  -  k  [A.  BJ  =  L, 

where  the  square  brackets  denote  the  concentration  of  the 
enclosed  ions  or  molecules,  the  former  in  gram-ions  per  litre, 
and  the  latter  in  gram-molecules  per  litre.  Since  the  solution 
is  in  equilibrium  with  solid  phase  the  concentration  of  the 
undissociated  part  [Am  BJ  for  a  given  temperature  is  constant 
and  independent  of  any  excess  of  either  of  the  ions  A-  or  B' 
in  the  form  of  another  electrolyte  with  a  common  ion.  The 
product 

[A-]-  [BT  -  L 

is  called  the  solubility-product  (or  ionic-product)  of  the  elec- 
trolyte. For  binary  electrolytes  (14)  takes  the  form 

(•5)  [A']  [B']  _  L, 

and  since  in  pure  aqueous  solution  [A*]  =  [B']  we  obtain  for 
that  case 


=  [B'j  = 

Further,  if  the  binary  electrolyte  is  very  insoluble,  and  be- 
longs to  the  class  of  strong  electrolytes,  like  AgCl,  for 
example,  we  may  regard  it  as  practically  completely  dissoci- 
ated in  solution,  and  the  concentration  of  the  undissociated 
part  as  negligible  compared  with  that  of  the  ions.  In  such  a 
case 


(16) 

where  S  is  the  solubility  of  the  electrolyte  in  gram-molecules 
per  litre. 

For  a  ternary  electrolyte,  e.g.  A2B,  we  obtain  similarly 

(17)  [A-]'  [B"]  =  L, 

and  since  in  pure  aqueous  solution  two  ions  of  A  are  formed 
for  one  of  B,  the  concentration  of  B"  in  gram  -ions  per  litre  is 
half  that  of  A-,  i.e.  [B"]  =  [A-J/2  or  [A-]  =  2  [B"J.  There- 
fore, from  (17), 


122          SOLUBILITY-PEODUCT—  EXAMPLES 


or 

(19)  4[B'?  [B"]  =  4[B"]»  =  L. 

Finally,  if  the  ternary  electrolyte  is  a  very  insoluble,  strong 
electrolyte,  so  that  practically  complete  dissociation  may  ba 
assumed,  from  (18)  or  (19)  the  solubility  S  of  the  electrolyte 
in  gram-molecules  per  litre  is 


since  one  molecule  of  the  undissociated  electrolyte  contains 
one  atom  of  B  and  2  atoms  of  A. 

Solubility-product  —  Examples 

PBOBLEM  240.—  At  20°  the  specific  conductivity  of  a  satu- 
rated solution  of  silver  bromide  was  1'576  x  10  ~  °  r.o. 
and  that  of  the  water  used  was  1-519  x  10  ~  6  r.o.  On  the 
assumption  that  the  AgBr  is  completely  dissociated,  calculate 
the  solubility  and  solubility-product  of  AgBr,  given  that  the 
'  equivalent  conductivities  of  KBr,  KNO3  and  AgNO3  at  infinite 
dilution  are  137*4,  131'3  and  121-0  r.o.  respectively. 

SOLUTION  240.  —  The  specific  conductivity  K  due  to  the 
AgBr  alone  is  (total  specific  conductivity  -  specific  conduc- 
tivity of  the  water  used), 

.-.  K  =  1-576  x  10  -6  -  1-519  x  10  ~6  =  0-057  x  10  ~6  r.o. 

Since  the  AgBr  is  assumed  to  be  completely  dissociated  the 
equivalent  conductivity  calculated  from  the  specific  conduc- 
tivity according  to  (4)  is  equal  to  the  equivalent  conductivity 
at  infinite  dilution  Aoo  .  A.  for  AgBr  may  be  obtained  from 
(8)  as  follows  :  — 

A«,    =  lAg.  +  ZBI,  =  (lAg.  +  ZN03')  +  (*K-+  JBT-)  -  fa-  +  W 
=  121-0  +  137-4  -  131-3 
=  127-1  r.o. 
Therefore,  from  (4), 

1000* 

A,  =  —_, 

127.1  ^  1000  x  0-057  x  10  -6 


SOLUBILITY-PKODUCT—  EXAMPLES          123 


1000  x  0-057  x  10  -  6 


- 


•  127-1 

where  c  is  the  concentration  of  the  saturated  solution,  or 
solubility,  in  gram-equivalents  per  litre. 

Since  the  AgBr  is  completely  dissociated,  the  concentration 
of  both  the  Ag  •-  and  Br'-ions  is  4*49  x  10  ~  7  gram-ion  per 
litre  and  the  solubility-product  is 

L  =  [Ag-][Br]  =  (4-49  x  10  ~7)2  =  2-03  x  lo'18. 

PROBLEM  241.  —  At  a.given  temperature  a  litre  of  a  saturated 
solution  of  silver  bromate  contains  S  =  0'0081  gram-molecule 
of  salt  ;  c  =  0-0085  gram-molecule  of  silver  nitrate  is  then 
added.  Calculate  the  new  solubility  S'  of  silver  bromate, 
assuming  that  both  salts  are  practically  completely  dissociated 
in  the  solution. 

SOLUTION  241.  —  In  the  saturated  pure  aqueous  solution  the 
concentration  of  both  the  Ag  •-  and  BrO3'-ions  is  S  gram-ion 
per  litre,  since  the  salt  is  completely  dissociated.  The 
solubility-product  of  AgBr03  is,  therefore, 

L  =  [Ag  -][BrO,']  =  (S)2  =  (0-Ocfel)2. 

In  any  solution  saturated  with  AgBrO3  the  product  of  the 
concentrations  of  the  Ag  •-  and  Br03'-ions  must  be  L.  Let 
S'  be  the  solubility,  in  gram-molecules  per  litre,  of  AgBrO3 
in  the  solution  containing  c  N-AgN03.  Since  both  salts 
are  assumed  to  be  completely  dissociated,  the  concentration 
of  the  Br03'-ions,  derived  from  the  dissociation  of  the  AgBrO3, 
is  S',  the  concentration  of  the  Ag  '-ions  from  the  AgBrO3  is  also 
S',  and  the  concentration  of  the  Ag  --ions  from  the  AgNO3  is 
c  gram  -ions  per  litre.  The  total  concentration  of  the  Ag  *-ions 
is,  therefore,  (c  +  S')  gram-ions  per  litre.  Therefore 

L  =  [Ag  -][Br03']  =  (c  +  S')(S'), 
or,  putting  in  the  numerical  values, 

(0-0085  +  S')(S')  =  (0-0081)2 
.'.  S'  =  0-0049  gram-molecule  per  litre. 

PROBLEM  242.—  The  solubility  of  Ag2C2O4  at  25°  is  S  = 
1-48  x  10  ~  4  gram-molecule  per  litre.  A  solution  of  potas- 
sium oxalate,  containing  c  =  0-2942  gram-  molecule  per  litre, 
was  shaken  at  25°  with  excess  of  Ag2CrO4  till  equilibrium 
was  established  according  to  the  equation 

Ag2Cr04  +  K2C204  =  Ag2C,04  +  K2Cr04. 


124          SOLUBILITY-PRODUCT— EXAMPLES 

The  solution  then  contained  c1  =  0-0602  .  gram  -  molecule 
K2CrO4  per  litre.  Assuming  that  the  degrees  of  dissociation 
of  the  oxalate  and  chromate  in  the  solution  are  equal  and 
that  the  dissociation  of  the  silver  salts  is  practically  complete, 
calculate  the  solubility-product  LCr  and  the  solubility  S'  of 
Ag2Cr04. 

SOLUTION  242. — The  solubility-product  of  Ag2C2O4  is 


£ra=[Ag-p[CA"] 

since,  on  complete  dissociation,  S  gram-molecule  of  Ag2C2O4 
gives  25  gram-ion  of  Ag  •  and  S  gram-ion  of  C204", 

.-.   Lox  =  4(1-48  x  10  -*)3  =  1-3  x  10 -11. 

In  the  mixed  solution  at  equilibrium  both  Ag2C2O4  and 
Ag2Cr04  are  present  as  solid  phases,  therefore  the  equations 

(1)  £c,r=  [Ag-H004"] 
and 

must  be  simultaneously  satisfied.  Since  the  Ag--ion  con- 
centration is  the  same  in  both,  namely  that  in  the  solution  in 
equilibrium  with  the  solid  phases,  we  obtain,  by  dividing  (1) 

by  (2), 

[C2O4"J  LOX 

The  concentration  of  the  K2CrO4  at  equilibrium  is  Cj  and 
of  the  K2C2O4  (c  -  Cj)  gram- molecules  per  litre.  If  a  is  the 
degree  of  dissociation  of  both  these  salts,  the  concentrations 
of  the  CrO4"-  and  C2O4"-ions  are  a.cl  and  a(c  —  cx)  gram-ions 
per  litre  respectively. 

From  (3),  therefore, 

[CrO4"]  =         acx 
[C204"]        a(c  -  Cj  " 
and 

c,  x  Ln9 


LCf 


0-0602  x  1-3  x  10 
0-2942  -  0-0602 


=  3  34  x  10 


-12 


If  S'  gram-molecule  per  litre  is  the^solubility  of  Ag2CrO4  in 
pure  aqueous  solution,  and  if  the  salt  is  completely  dissoci- 


SOLUBILITY-PRODUCT—EXAMPLES          125 

ated,   the  concentration   of  the  Ag  '-ions  is  2S'  and  of  the 
CrO4"-ions  S'  gram-ions  per  litre.     Therefore 


[Ag-p  [CKY]  =  (2S7  (Sf)  =  4(57  =  L 
and 


10  "12 

—     -  =  0-94  x  io~4  gram- 


molecule  per  litre. 

PROBLEM  243.  —  The  dissociation-constant  of  ammonium 
hydroxide  is  K  =  1'8  x  10  ~5,  and  the  solubility-product  of 
magnesium  hydroxide  is  L  =  1'22  x  10  ~n.  How  many 
grams  of  solid  ammonium  chloride  must  be  added  to  a  mixture 
of  50  c.c.  of  N-NH4OH  solution  and  50  c.c.  of  N-MgCl2 
solution,  so  that  the  precipitate  of  magnesium  hydroxide 
may  just  disappear  ?  Assume  that  the  volume  of  the  solution 
is  not  changed  by  dissolving  the  solid  NH4C1,  and  that  the 
dissociation  of  the  neutral  salts  is  complete. 

SOLUTION  243.  —  In  any  solution  saturated  with  Mg(OH)2 
the  equation 


must  be  satisfied.  The  precipitate  of  Mg(OH)2  disappears, 
therefore,  in  a  solution  in  which  the  concentration  of  the  OH'- 
ions  is  given  by  the  equation 


t°H'J  •  VfHfT 


where  [Mg  •  •]  is  equal  to  the  total  amount  of  magnesium 
present.  The  equilibrium  equation  for  the  electrolytic  dis- 
sociation of  the  ammonia  in  the  solution,  namely, 

[NH4-]  [OH']  =  K  [NH4OH], 

must  also  be  satisfied. 

The  NH4  --ions  are  derived  practically  entirely  from  the 
solution  of  the  solid  ammonium  chloride,  since  the  NH4OH 
can  be  regarded  as  practically  undissociated  in  presence  of 
the  NH4C1.  The  concentration  of  the  NH4--ions,  is,  there- 
fore, 


[OH'] 


126          SOLUBILITY-PRODUCT—EXAMPLES 

In  order  to  produce  this  concentration  of  NH4'-ions  we  must 
dissolve  in  the  100  c.c.  of  solution  (  =  1/10  litre) 


x  =  .£[NH4OH]  x        grams  of  NH4C1, 

where  M  denotes  the  molecular  weight  of  NH4C1. 

Substituting  the   numerical    values   in   this   equation   we 
obtain 


1-8  x  10-  x  0-5  x  °'5 


x10-u 
=  9-8  grams  NH4C1, 

since  the  total  concentration  of  both  NH4OH  and  Mg  in  the 
mixed  solutions  is  0*5  N. 

PROBLEM  244.  —  The  solubility  of  calcium  carbonate  in  pure 
water  is  S  =  1-3  x  10  ~4  gram-molecules  per  litre.  What  is 
its  solubility  in  water  which  is  saturated  with  carbon  dioxide 
under  a  pressure  (1)  of  _pj  =  ^  atmosphere,  and  (2)  of  p2 
atmosphere,  and  which,  therefore,  contains  carbonic  acid,  i 
the  concentration  of  carbonic  acid  in  water  is  K^  =  0'04354 
times  the  pressure  of  the  carbon  dioxide  in  atmospheres,  and 
the  first  and  second  dissociation-constants  of  carbonic  acid  in 
water  are  K2  =  3-04  x  10  ~7  and  Ks  =  1-3  x  10  ~14  respec- 
tively ? 

SOLUTION  244.  —  In  the  case  of  all  solutions  which  are 
saturated  with  CaCO3  the  equation  [Ca  •  •]  [CO8"]  =  L  must 
be  satisfied,  where  L  is  the  solubility-product  of  calcium  car- 
bonate. In  pure  aqueous  solutions  we  may  assume  that  the 
ionic  dissociation  is  complete,  and  can,  therefore,  put  [Ca  •  •] 
=  [CO,"]  =  S.  We  thus  obtain 

(1)  [Ca--][C03"]  =  L  =  S*. 

By  the  addition  of  free  carbonic  acid  the  solubility  of  calcium 
carbonate  is  increased,  i.e.  the  concentration  of  the  Ca  •  •- 
ions  is  increased  owing  to  the  diminution  of  the  CO3"-ion 
concentration  by  the  reaction 

CO3"  +  H2CO3  =  2HC03', 

or,  in  other  words,  owing  to  the  ,  formation  of  bicarbonate. 
The  mass-action  equation  for  this  reaction  is 

~ 


[H.CIOJ 


SOLUBILITY-PRODUCT—  EXAMPLES          127 

This  constant  K  may  be  calculated  as  follows,  from  the  given 
values  of  the  dissociation-constants  of  carbonic  acid.  For  the 
first  dissociation  of  carbonic  according  to  the  equation  H2CO3 
=  H-  -f  HCO3',  the  law  of  mass-action  gives 

[H-J  [H00.1  _  K 
-[3J50T 

and  for  the  second  dissociation  according  to  the  equation 
HCO3'  =  H-  +  CO3"  we  obtain  similarly 

H-nccv] 


(3)  divided  by  (4)  gives  (2),  thus 

[HOO.T  _i, 

[00,"]  [  ' 


By  multiplying  (1)  by  (2)  we  obtain 

[Oa--][HCO,T  „  «  ,  J& 
[H.OOJ  '  K- 

The  law  of  electro-neutrality  requires 

2[Ca--]  +  [H-]  =  [HC08']  +  2[CO3"]. 

Since  carbonic  acid  is  a  very  weak  acid,  [H¥]  is  very  small 
even  in  presence  of  free  H2C03,  and  [C03"]  may  be  neglected 
in  comparison  with  [HG03'].  We  thus  obtain 

2[Ca  •  •]  =  [HC03']. 

The  concentration  of  Ca  •  --ions,  and,  therefore,  the  solu- 
bility of  calcium  carbonate'  in  water  containing  carbonic  acid, 
may  be  calculated  from  equation  (5)  if  the  concentration  of 
the  free  carbonic  acid,  [H2CO3],  in  the  solution  is  known. 
This  may  be  obtained  with  the  help  of  the  factor  of  propor- 
tionality Klt  which  is  given.  If  the  partial  pressure  of  the 
CO2  over  the  solution  is  p,  then 

[H.2C03]  =  KlP. 

Substituting  this  value  of  [H2C03]  in  equation  (5),  and 
bearing  in  mind  that  2[Ca-  •]  =  [HC03'],  we  obtain  from  (5), 

' 


and 


128  HYDKOLYSIS 


3  /(l-3)2x  10 -8x  4-354  x  10 -2  x  3-04  x  10 - t 
=  V  4  x  1'3  xlO~n  "  ^ 

0-016  55". 
^5  atmosphere 
[Ca  ' ']  =  0*016  x  0*37  =  0*0059  gram -molecule  per  litre, 
and  for  p2  =  %  atmosphere 
[Ca  ' ']  —  0'016  x  0-794  =  o'oi27  gram-molecule  per  litre. 

Hydrolysis  of  Salts 

In  an  aqueous  solution  of  a  salt  BA  of  a  weak  acid  HA 
and  a  strong  base  BOH  the  salt  is  more  or  less  hydrolysed 
according  to  the  equation 

BA  +  H2O  =  HA  +  BOH. 

Let  x  be  the  degree  of  hydrolysis,  that  is,  of  each  gram- 
molecule  of  salt  in  solution  let  the  fraction  x  be  hydrolysed 
according  to  the  above  equation  ;  the  fraction  (1  -  x)  of  each 
gram-molecule  remains,  therefore,  unhydrolysed.  If  G  is  the 
total  concentration  of  the  salt  in  gram-molecules  per  litre, 
the  concentration  of  HA  and  of  BOH  is,  therefore,  xc,  since 
they  are  produced  in  equivalent  amounts,  and  the  concentra- 
tion of  the  unhydrolysed  salt  BA  is  (1  -  x)c.  We  shall 
assume  further  that  the  strong  electrolytes,  the  salt  BA  and 
the  base  BOH,  are  completely  dissociated,  whilst  the  weak 
electrolyte,  the  acid  HA,  is  practically  undissociated,  a  condi- 
tion which  will  be  very  approximately  fulfilled,  since  the  dis- 
sociation of  the  weak  acid  into  H*-  and  A'-ions,  which  is  feeble 
even  in  a  pure  aqueous  solution,  will  be  still  further  diminished 
by  the  presence  of  the  large  excess  of  A'-ions  derived  from  the 
dissociation  of  the  salt  BA.  If  the  concentrations  of  the 
various  molecules  and  ions  are  denoted  by  square  brackets, 
we  may,  therefore,  put 

[HA]  =  xc,  [OH']  =  xc  a-nd  [A']  =  (1  -  x)c, 

since,  in  the  last  case,  the  concentration  of  the  A'-ions  derived 
from  the  dissociation  of  the  HA  must,  for  the  reasons  given 
abjjre,  be  practically  negligible.  The  concentration  of  the 
Herons,  [H*],  we  obtain  from  the  fact  that  in  any  aqueous 


HYDROLYSIS  129 

solution  at  a  given  temperature  the  product  of  the  concen- 
trations of  the  IT-  and  OH'-ions  must  be  constant,  or 


Kw  is  called  the  ionic  product  of  water.     We  obtain,  there- 
fore 


• 
[OH']        xc 

Applying  the  law  of  mass-action  to  the  dissociation  of  the 
weak  acid  HA  we  obtain 

-T7- 

_.    [H-][A']_^  -  K 

~  ^ 


where  Ka  is  the  dissociation-constant  of  the  acid.     Hence 


Kf!  is  called  the  hydrolysis-constant  of  the  salt  BA  and,  as 
is  evident  from  equation  ^21), 

-g    =  (cone,  of  free  acid)(conc.  of  free  base)  ^ 
(cone,  of  unhydrolysed  salt) 

If  Kft  and,  therefore,  the  degree  of  hydrolysis  x,  is  very 
small,  that  is,  if  Ka  is  very  much  greater  than  Kw,  we  may 
put  (1  -  x)  =  1  without  great"  error,  and  we  obtain  from  (21) 
the  approximation  formula 

(22)  K,  =  x*c. 

In  an  exactly  similar  way  it  may  be  shown  that  for  a  salt 
BA  of  a  weak  base  BOH  and  a  strong  acid  HA, 

(•**}  K    -  K«  -     x*c 
(23)  Km  -  ^  -  -_, 

where  KH  is  the  hydrolysis-constant  of  the  salt,  Kb  the  dissoci- 
ation-constant of  the  weak  base,  x  the  degree  of  hydrolysis 
of  the  salt  and  c  the  total  concentration  of  the  salt. 

If  the  salt  BA  of  the  weak  acid  HA  and  the  weak  base 
BOH  is  hydrolysed  according  to  the  equation 

BA  +  H2O  =  HA  +  BOH 

and  if  we  assume  that  the  dissociation  of  the  unhydrolysed 
salt  (a  strong  electrolyte)  is  complete,  whilst  the  weak  acid 
9 


130  HYDROLYSIS 

and  base  are  practically  undissociated  in  presence  of  the 
excess  of  A  •  and  B'-ions  furnished  by  the  dissociation  of  the 
salt,  then  if  c  molecules  per  litre  is  the  total  concentration  of 
BA,  and  x  is  the  degree  of  hydrolysis, 

[A-]  =  (1  -  x)c,  [B'J  =  (1  -  x)c,  [HA]  =  xc,  [BOH]  =  xc, 

where  the  square  brackets  denote  the  concentrations  of  the 
enclosed  substances  in  gram-molecules  per  litre.  Therefore 


"  [HA]  xc 

']_(1  -qQc[OH'] 


[BOH]  xc 

and 

Z    x  K  =  [H  "I 


XC  XC 


Therefore 

It 


(24) 


x 


Kh 


_  (cone,  of  free  acid) (cone,  of  free  base) 
(cone,  of  unhy Jrolysed  salt)'2 

Since  the  concentration  c  does  not  appear  in  this  equation 
the  degree  of  hydrolysis  of  a  salt  of  a  weak  acid  and  a  weak 
base  is  independent  of  the  concentration. 

Hydrolysis— Examples 

PBOBLEM  245. — At  25°  C.  the  dissociation-constant  of 
ammonia  in  aqueous  solution  is  Kb  =  1-8  x  10  ~5,  and  at  the 
same  temperature  the  ionic  product  of  water  is  Kw  =  0'8  x  10  ~ 14. 
The  concentration  of  hydrogen  ions  in  a  solution,  which 
is  cl  =  0-01  N  with  respect  to  ammonia  and  c2  =  0-02  N 
with  respect  to  diketotetrahydrothiazole,  is  a  =  T7  x  10  ~7 
gram-ion  per  litre.  What  is  the  dissociation-constant  Ka  of 
the  acid,  diketotetrahydrothiazole,  and  to  what  degree  per 
cent,  is  the  neutral  ammonium  salt  of  the  acid  hydrolysed 
(1)  inac3  =  0-001  N-solution,  and  (2)  in  a  C4  =  O'l  N-solution  ? 

SOLUTION  245. — The  acid,  diketotetrahydrothiazole,  may, 


HYDEOLYSIS—  EXAMPLES  131 

for  brevity,  be  represented  by  the  formula  AH,  in  which  H  is 
the  acid  hydrogen  or  cation,  and  A  the  acid  radical  or  anion. 
According  to  the  law  of  mass-action,  the  equilibrium 
equations  for  the  dissociation  of  the  base,  NH4OH,  and  of 
the  acid,,-  AH,  are 

[NH4-]  [OH']  _ 
*  '      [NH4OH]  6  . 

and 


For  the  dissociation  of  water  the  ionic  product  is 
(3)  [H-]  [OH']  =  Kw. 

In  an  aqueous  solution  containing  both  the  base  and  the 
acid,  all  three  equations  must  be  simultaneously  satisfied. 
In  these  equations  Ka  and  the  concentrations  [OH'],  [NH4'], 
[NH4OH],  [A']  and  [AH]  are  unknown,  whilst  [H*]  is  given 
=  a.  For  the  evaluation  of  these  six  unknowns  three 
further  equations  are,  therefore,  required.  These  are 

(4)  [NH4-]  +  [NH4OH]  =  cv 

(5)  [A']  +  [AH]  =  c2 

and  (6)  [NH4-]  +  [H-]  =  [A']  +  [OH']. 

The  concentration  of  the  undissociated  salt  molecules, 
[NH4A],  may  be  neglected,  since  the  salt  may  be  regarded 
as  practically  completely  dissociated.  The  problem  is,  there- 
fore, capable  of  solution,  and  its  solution  is  simplified  by  the 
fact  that  [H-]  =  a  is  very  small  compared  with  cl  and  c>2  ; 
the  anions  A'  are,  therefore,  derived  practically  entirely  from 
the  dissociation  of  the  salt  NH4A,  and  the  the  excess  of  the 
acid  AH  is  only  slightly  dissociated. 

Equation  (6)  therefore  simplifies  to 

(7)  [NH4-]  =  [A']. 
From  (1)  and  (3)  we  obtain 

[NHJ  Kb         Kh{R-]  _  1-8  x  10-5  x  jL-7  x  ip-7 

[NH4OH]      [OH']         K~  0-8  xTO"-'1* 

382 
"I" 

[NH4OH]  may,  therefore,  be  neglected  in  comparison  with 
[NH4-].  We  thus  obtain 


132  HYDEOLYSIS— EXAMPLES 

from  (7)  and  (4)  :  [NH4-]  =  [A']  =  cl  =  0-01, 
from  (5) :  [AH]  =  c.2  -  c1  =  0-01, 


and  from(2)  :  Ka  =  "Tn^fH  =  [H-]  =  a  =  1-7  x  IO-7. 


In  a  c3  N-solution  of  the  neutral  ammonium  salt  let  the 
fraction  x  be  hydrolysed  according  to  the  equation 

NH4A  +  H20  =  AH  +  NH4OH. 
Then  [NH4OH]  =  [AH]  =  xcz, 
and  [NH4-]  =  [A'J  =  c3  -  xc,. 

The  ions  H-  and  OH'  are  present  only  in  very  small  con- 
centration, since  both  the  base  and  the  acid  are  very  weak  ; 
and  the  concentration  of  the  undissociated  salt  molecules, 
[NH4A],  may  again  be  neglected  on  account  of  the  practically 
complete  dissociation  of  the  salt. 

From  (1)  we  thus  obtain 

(8) 
and  from  (2) 


(8)  x  (9)  gives 

n      ^2 

\L  ~  ^)    rmrn  nr.n        v 
xz       LOH  J  LH  ]  =  Kb 

and  by  dividing  this  result  by  (3)  we  obtain 


and 


iKJta      1  +     /1-8  x  1-7 

\  J^  V     0-8  x  : 


-J       1+  x/382 


=  0-049. 

The  degree  of  hydrolysis  is,  therefore,  4-9  per  cent.,  and  is 
the  same  for  all  concentrations  of  the  neutral  salt,  since  x  is 
independent  of  the  total  concentration  cy  The  degree  of 


HYDEOLYSIS— EXAMPLES  133 

hydrolysis   of  a   salt  of  a  weak  acid   and  a  weak  base  is, 
therefore,  independent  of  the  concentration. 

PROBLEM  246. — At  25°  the  distribution- coefficient  of  ani- 
line between  benzene  and  water  is  10*1.  A  litre  of  0'03138 
N-aniline  hydrochloride  was  shaken  with  59  c.c.  of  benzene. 
After  equilibrium  was  established  50  c.c.  of  the  benzene  was 
found  to  contain  0-02916  gram  of  aniline.  Calculate  (a)  the 
hydrolysis-constant  of  aniline  hydrochloride,  (6)  the  percent- 
age hydrolysis  in  0*1  N-solution  and  (c)  the  dissociation- 
constant  of  aniline  as  a  base,  given  that  the  ionic-product  for 
water  at  25°  is  1-2  x  10  -14. 

SOLUTION  246. — (a)  The  molecular  weight  of  aniline  is  93, 
therefore 

(1)  50  c.c.  of  benzene  contain  — — —  gram- equivalent  of 

93 

aniline,  and  the  concentration  of  the  aniline  in  benzene  is 

2^  x  1000  gram.equivalent  per  litre. 
93  oO 

Since 

concentration  of  aniline  in  benzene  __  JQ.-^ 

concentration  of  aniline  in  water 
we  obtain 

(2)  concentration  of  free  aniline  in  water  layer 

concentration  in  benzene  _  0-02916  x  1000 
10-1  "  93  x  50  x  10-1 

=  0-000621  gram-equiv.  per  litre. 

The  total  concentration  of  aniline  originally  present  in  the 
water  was  0*03138  gram-equiv.  per  litre,  but,  from  (1),  of  this 
amount  59  c.c.  of  benzene  extracted 

0*02Qlfi        ^Q 

— x   _  =  0-00037  gram-equiv. 

93  50 

The  total  concentration  of  aniline   (free   and   combined   as 

hydrochloride)  in  the  water  layer  at  equilibrium  is,  therefore, 

0-03138  -  0-00037  =  0-03101  gram-equiv.  per  litre. 

From  (2),  however,  the  concentration  of  the  free  aniline  is 
0-000621  gram-equiv.  per  litre,  therefore  the  concentration  of 
the  aniline  hydrochloride  is 

(3)  0-03101  -  0-000621  =  0-03039  gram-equiv.  per  litre. 


134  FAEADAY'S  LAWS—  PEOBLEMS 

Since  the  benzene  extracts  only  aniline  and  not  hydrochloric 
acid  or  aniline  hydrochloride  from  the  aqueous  layer,  the  total 
concentration  of  HC1  in  the  aqueous  layer  at  equilibrium  is 
equal  to  that  of  the  aniline  hydrochloride  originally  present, 
namely  0*03138  gram  equiv.  per  litre.  Of  this  amount,  how- 
ever, 0-03039  gram-equiv.  exists  as  aniline  hydrochloride 
(from  (3)),  therefore  the  concentration  of  free  HC1  at  equi- 
librium is 

0-03138  -  0-03039  =  0-00099  gram-equiv.  per  litre. 
.From  (21) 

-g    _  (cone,  of  free  acid)  (cone,  of  free  base) 

(cone,  of  unhydrolysed  salt) 
_  (0-00099)(0-000621)  . 

'  (0-03039) 
(b)  If  x  is  the  degree  of  hydrolysis  in  O'l  N-solution,  from 

(21) 

KH  =  -*----  or  2-02  x  10-  5= 


1  -  x  1  -  x 

.'.  x  -  1-41  x  lO-2 
=  1^41  per  cent. 

(o)  From  (21)  KH  = 


Problems  for  Solution 

In  the  following  problems  the  answers  to  those  marked  * 
have  been  obtained  by  using  the  simplified  formula?  (12)  and 
(13)1  i-G.  by  putting  (1  -  a)  =  1,  and  (22),  i.e.  by  putting 

(i  -  *)  -  1. 

Faraday's  Laws 

PROBLEM  247.  —  A  current  passed  through  a  water  volta- 
meter liberated  in  4  minutes  50  c.c.  of  hydrogen  at  17°  C. 
and  750  mm.  Calculate  the  mean  current  during  the  whole 
time. 

Ans.  1'669  amp. 

PROBLEM  248.  —  In  the  preparation  of  NaOH  by  the  electro- 
lysis of  a  sodium  chloride  solution,  600  c.c.  of  solution  con- 
taining 40  grams  NaOH  per  litre  was  obtained  after  a  certain 


TEANSPORT  NUMBERS— PROBLEMS         135 

time.  During  the  same  time  3O4  grams  of  Cu  had  been 
deposited  in  a  copper  voltameter  in  series  with  the  electro- 
lytic cell.  Calculate  the  percentage  of  the  theoretical  yield 
of  NaOH  obtained. 

Ans.  62'8  per  cent. 

PROBLEM  249. — What  volume  of  hydrogen  at  18°  and  737 
mm.  is  liberated  by  passing  a  current  of  1-54  arnp.  through 
a  solution  of  dilute  H2SO4  for  2  minutes  ? 

Ans.  23-56  c.c. 

PROBLEM  250. — A  current  of  1-5  amp.  is  passed  through 
a  solution  of  CuCl2  for  1  hour.  What  weight  of  electrolyte 
is  decomposed  ? 

Ans.  3*764  grams. 

Transport  Numbers 

PROBLEM  251. — A  solution  of  AgN03  containing  1-139  mg. 
of  silver  per  gram  of  solution  was  electrolysed  between  silver  \ 
electrodes  and  the  anode  liquid  after  electrolysis  contained 
39-66  mg.  silver  in  2O09  grams  of  solution.  In  a  silver  volta- 
meter  in  series  with  the  electrolytic  cell  32' 10  mg.  of  silver 
was  deposited.  Calculate  the  transport  numbers  of  Ag-  and 
N0'3. 

Ans.  Ag  •  =  0/476,  NO'3  -  0-524. 

PROBLEM  252. —  A  solution  of  CuSO4  containing  1  gram 
CuSO4  per  41-59  grams  water  was  electrolysed  between  a 
copper  anode  and  a  platinum  cathode,  a  silver  voltameter 
being  placed  in  series  with  the  electrolytic  cell.  After  electro- 
lysis 54-706  grams  of  the  cathode  solution  gave  on  analysis 
0-5118  gram  CuO.  During  the  electrolysis  0*6934  gram  of 
silver  was  deposited  in  the  voltameter.  Calculate  the  trans- 
port numbers  of  S0"4  and  Cu  •  •.  (Cu  =  63-6,  S  =  32-06, 
O  =  16,  Ag  =  107-9.) 

Ans.  SO"4  =  0-5146,  Cu  •  •  =  0-.4854. 

PROBLEM  253. — A  solution  of  NaCl  containing  0*01784  per 
cent,  of  chlorine  was  electrolysed  between  a  Cd  anode  and  a 
Pt  cathode,  with  a  silver  voltameter  in  series.  After  electro- 
lysis the  solution  was  divided  into  three  portions,  cathode, 
anode,  and  middle  portion.  The  concentration  of  the  latter 
was  unchanged,  whilst  226-99  grams  of  anode  solution  con- 
tained 0-04679  gram  chlorine,  and  331-49  grams  cathode 
solution  contained  0 '05302  gram  chlorine.  In  the  silver 


v/v. 

E 


136     ELECTROLYTIC  DISSOCIATION— PROBLEMS 

voltameter  the  equivalent  of  0*01021  gram  chlorine  was  de- 
posited.    Calculate  the  transport  numbers  of  Cl'  and  Na\ 

Ans.  Cl'  =  0-611,  Na-  =  0*389. 

PROBLEM  254. — A  solution  of  CdCl2  containing  0-2016  per 
cent,  of  chlorine  was  electrolysed  between  a  Cd  anode  and  a 
t  cathode.     After   electrolysis   33-59   grams   of   the  anode 
quid  contained  0-08020   gram  Cl  and  54*12  grams  of  the 
athode  liquid  contained  0-09662  gram  Cl.     The  concentra- 
tion of  the  middle  portion  was  unchanged.     In  a  silver  volta- 
meter in  series  with  the  electrolytic  cell  0-06662   gram  of 
silver  was  deposited  during  the  electrolysis.     Calculate  the 
transport  numbers  of  Cl'  and  Cd  •  •. 

Ans.  01'  -  0-570,  Cd  •  •  -  0  430. 

Electrolytic  Dissociation 

PROBLEM  255. — The  equiv.  conductivity  of  LiCl  at  18°  is 
101-4  r.o.  at  infinite  dilution  and  93-6  r.o.  for  a  O'Ol  N-solution. 
What  is  the  degree  of  dissociation  and  the  concentration  of 
the  Cl'  ions  in  this  solution  ? 

Ans.  a  =  0-923,  [01']  =  0-00923  gram-ion/litre. 

PROBLEM  256. — At  18°  the  equiv.  conductivity  of  HI  at 
infinite  dilution  is  384  r.o.  and  the  specific  conductivity  of  a 
0-405  N-solution  is  0-1332  r.o.  What  is  the  concentration  of 
the  H  --ions  in  this  solution  ? 

Ans.  [H  •]  =  0-347  gram-ion/litre. 

PROBLEM  257. — The  equiv.  conductivity  at  25°  of  acetic  acid 
containing  1  gram-equiv.  in  32  litres  is  9'2  r.o.  The  equjv. 
conductivity  at  infinite  dilution  is  389  r.o.  Find  the  dissoci- 
ation-constant of  the  acid. 

Ans.  1-79  x  10-6. 

PROBLEM  258. — At  25°  the  specific  conductivity  of  butyric 
acid  at  a  dilution  of  64  litres  is  1-812  x  10  ~  4  r.o.  The  equiv. 
conductivity  at  infinite  dilution  is  380  r.o.  What  is  die 
degree  of  dissociation  and  the  concentration  of  H  --ions  in^he 
solution,  and  the  dissociation-constant  of  the  acid  ? 

Ans.  a  =  0-0305,  [H  •]  =  4 -765  x  10  -  4  gram-ions/litre, 
K=  1-5  x  10 -5. 

PROBLEM  259  (cf.  preceding  problem). — What  is  the  value 
of  the  dissociation-constant  of  butyric  acid  if  the  concentra- 


ELECTROLYTIC  DISSOCIATION—  PROBLEMS     137 


tion  is  measured  in  gram-equivalents  per  c.c.  instead 
equivalents  per  litre  ? 

Ans.  1-5  x  10  ~8. 

PROBLEM  260.  —  The  specific  conductivity  of  a  5  per  cent. 
(by  weight)  BaCl2  solution  is  389  x  10  ~  4  r.o.  and  its  density 
1-0445  at  18°.  What  is  the  equivalent  conductivity  and  de- 
gree of  dissociation  of  the  solution,  if  the  equiv.  conductivity 
of  BaCl2  at  infinite  dilution  is  123  r.o.  at  18°  ? 
Ans.  A  =  77-7,  a  =  0-632. 

PROBLEM  261  (cf.  preceding  problem).  —  What  is  the  freez- 
ing-point of  the  BaCl2  solution  ? 

Ans.     -  1-064°. 

PROBLEM  262.  —  At  25°  the  specific  conductivity  of  ammonia 
solutions,  containing  c  gram-equivs.  per  litre  is  *  r.o.  The 
ionic  conductivity  of  the  NH4'-ion  at  this  temperature  is  70-4 
and  of  the  OH'-ion  200-6.  Calculate  the  equiv.  conductivity 
and  degree  of  dissociation  at  each  concentration  and  the  mean 
dissociation-constant. 

c  •        K 

0-0109  1-220  x  10-' 

^0-0219  1-730  x  10  ~4 

Ans.    A  =  11-2  and  7'9  r.o.,  a  =  0-0413  and  0-0231, 
K  =  1-94  x  10~5and  1-92  x  10  ~5,  mean  =  1-93  x  10  ~5. 

*  PROBLEM  263  (cf.  preceding  problem).  —  At   what  con- 
centration is  ammonia  1  per  cent,  dissociated  in  solution  ? 
Ans.   c  =  0-193  N. 

PROBLEM  264.  —  At  25°  the  specific  conductivity  of  ethyl- 
amine at  a  dilution  of  16  litres  is  1-312  x  10  ~  3  r.o.,  the  equiv. 
conductivity  at  infinite  dilution  is  232-6  r.o.  Calculate  the 
dissociation-constant. 

Ans.   5-6  x  lO-4. 

PROBLEM  265  (cf.  preceding  problem).  —  What  is  the  con- 
centration of  OH'-ions  in  the  ethylamine  solution  ? 
Ans.    5-64  x  10  ~  3  gram  -ion/litre. 

PROBLEM  266.  —  At  what  concentration  of  ethylamine  is  the 
concentration  of  the  OH'-ions  0-01  N  ? 

Ans.    c  =  0-1886  mole/litre. 
PROBLEM  267.  —  At  25°  the  dissociation-constant  of  mono- 


138    ELECTROLYTIC  DISSOCIATION— PROBLEMS 

chloracetic  acid  is  T55  x  10  ~3,  and  its  equiv.  conductivity 
at  a  dilution  of  32  litres  is  77-2  r.o.  What  is  its  equiv. 
conductivity  at  infinite  dilution  ? 

Ans.    388  r.o. 

*  PROBLEM  268. — At  25°  the  dissociation-constant  of  lactic 
acid  is  1*4  x  10  ~4  and  of  monochloracetic  acid  1-55  x  10  ~3. 
At  what  concentrations  of  lactic  and  chloracetic  acids  is   the 
H  --ion  concentration  =  O'Ol  N  ? 

Ans.   For  lactic,  c  =  0-724  N. 

For  chloracetic,  c  =  0-0745  N. 

PROBLEM  269. — The  specific  conductivity  of  KC1  is  practi- 
cally proportional  to  its  concentration  in  solutions  of  moderate 
concentration.  The  specific  conductivity  at  18°  of  a  10  per 
cent.  KC1  solution  is  01359  r.o.  and  of  a  15  per  cent,  solution 
0-2020  r.o.  What  is  the  percentage  concentration  of  a  KC1 
solution  of  which  the  specific  conductivity  is  0-1640  r.o.  ? 
Ans.  12-13  per  cent. 

PROBLEM  270.— The  density  at  0°  of  a  Ca(NO3)2  solution 
containing  0-208  gram -molecule  per  litre  is  1*010.  Its  freez- 
ing-point is  —  0'910°  and  its  molecular  conductivity  at  0°  is 
78-8  r.o.  Taking  the  molecular  conductivity  at  infinite  dilu- 
tion =  129  r.o.  at  0°  and  the  freezing-point  constant  for  water 
=  1-86  (gram -mole  in  1000  grams  water),  calculate  the  degree 
of  dissociation  from  the  conductivity  and  from  the  freezing- 
point. 

Ans.    From  conductivity     =  0-611. 
From  freezing-point  =  0-648. 

PROBLEM  271. — At  25°  the  specific  conductivities  of  KC1 
and  LiCl  at  a  dilution  of  32  litres  are  4-25  x  10  ~ 3  r.o.  and 
3-243  x  10  ~ 3  r.o.  respectively.  Compare  the  degrees  of 
dissociation  of  the  two  salts  at  this  dilution,  taking  the  ionic 
conductivities  of  K,  Li  and  Cl  at  25°  as  75'3,  41  and  76  r.o. 
respectively. 

Ans.   KC1  =  0-899,  LiCl  =  0*892. 

*  PROBLEM  272. — The  dissociation-constant  of  acetic  acid 
at  18°  is  1*8  x  10  ~5.     Calculate  the  degree  of  dissociation 
and  the  H'-ion  concentration  (1)  in  a  0-25  N-acetic  acid  solu- 
tion, (2)  in  a  0*25  N-acetic  acid  solution  containing  0'25  N- 
sodium  acetate,  if  the  sodium  acetate  is  assumed  to  be  com- 
pletely dissociated. 


ELECTKOLYTIC  DISSOCIATION— PROBLEMS  '139 

An3.(l)a  =  848  x  10-3,  [H«]  =  2'12  x  10  - 3  gram-ion/litre, 
(2)  a  =  7  x  10  ~4,  [H-]  =  1-75  x  10    4  gram-ion/litre. 

PROBLEM  273. — At  18°  the  specific  conductivity  of  a  5  per 
cent,  solution  of  Mg(NO3)2  is  438  x  10  ~4  r.o.  and  its  density 
1-0378.     What  is  its  degree  of  dissociation  if  the  equivalent 
conductivity  of  Mg(NO3)2  at  infinite  dilution  is  109 '8  r.o.  ? 
^Ans.  0-57. 

PROBLEM  274. — At  18°  the  molecular  conductivity  of 
LiNOa  is  94-46  r.o.  at  infinite  dilution,  and  75'01  r.o.  in  a 
0*2  N-solution.  What  is  the  concentration  of  Li  '-ions  in 
this  solution  ? 

Ans.  0-159  N. 

PROBLEM  275. — The  density  of  a  5  per  cent.  NaCl  solution 
at  18°  is  1-0345  and  its  specific  conductivity  672  x  10  ~4  r.o. 
What  is  (1)  the  molecular  conductivity,  (2)  the  degree  of 
dissociation  of  the  solution,  if  the  molecular  conductivity  of 
NaCl  at  infinite  dilution  is  109  r.o.  ?  (3)  What  is  the  vapour- 
pressure  of  the  solution  at  18°  if  that  of  pure  water  is  15*33 
mm.  ? 

Ans.  (1)  76  r.o.,  (2)  0-697,  (3)  14-91  mm. 

PROBLEM  276. — The  velocity  of  migration  of  the  Ag--ion 
at  18°  is  0-000577  cms.  per  sec.  and  of  the  NO'3-km  0-000630 
cms.  per  sec.  The  specific  conductivity  of  0-1  N-AgN03  at 
18°  is  0-00947  r.o.  What  is  the  degree  of  dissociation  of  the 
AgN03? 

Ans.  81-3  per  cent. 

PROBLEM  277. — The  freezing-point  of  a  0*1  molecular 
N  -  CaCl.,  solution  is  -  0-482°.  (1)  Calculate  the  degree  of  dis- 
sociation (freezing-point  constant  =  1-89  for  gram-mol.  per 
litre).  (2)  Compare  the  value  with  that  found  from  the  equi- 
valent conductivity  at  18°,  which  is  82'79  r.o.,  whilst  the 
equivalent  conductivity  of  CaCl2  at  infinite  dilution  is  115*8 
r.o. 

Ans.  (1)  0-775,  (2)  0-715. 

*  PROBLEM  278. — What  is  the  concentration  of  H'-ions  in 
a  solution  containing  1  gram-molecule  of  acetic  acid  and  1 
gram-molecule  of  cyanacetic  acid  per  litre  ?  The  dissociation- 
constant  of  acetic  acid  is  18  x  10  ~ 5  and  of  cyanacetic  acid 
370  x  10 -5. 

Ans.  6-1  x  10  -*N. 


140     ELECTEOLYTIC  DISSOCIATION— PEOBLEMS 

PROBLEM  279. — The  molecular  conductivity  at  25°  of  ethyl 
hydrogen  malonate  is  356  at  infinite  dilution,  and  21*5,  41*9 
and  57*3  at  the  dilutions  8-57,  34-28  and  68-56  litres  respec- 
tively. Calculate  the  degree  of  dissociation  and  the  dissocia- 
tion-constant at  each  dilution. 

Ans.    a  =  6-04,  11-8  and  16-1  per  cent. 

K  =  4-54  x  10  - 4,  4-58  x  10  ~  4  and  4-51  x  10  - 4. 

PROBLEM  280. — At  25°  the  molecular  conductivities  of 
malonic  acid  at  the  dilutions  32,  64  and  128  litres  are  77' 1, 
103-6  and  137'0  r.o.  respectively.  The  molecular  conductivity 
at  infinite  dilution  for  dissociation  into  IT-  and  C3H30'4-ions 
is  382  r.o.  Calculate  the  degree  of  dissociation  and  the  dis- 
sociation-constant at  each  dilution.  Whaf;  conclusion  may  be 
drawn  from  these  figures  as  to  the  manner  in  which  malonic 
acid  dissociates  at  these  dilutions  ? 

Ans.    a  =  0-202,  0  271,  0-359 

K=  1-59  x  10  ~3,     1-58  x  10  ^3,     1-57  x  10  ~3. 

PROBLEM  281  (of.  preceding  problem). — For  the  dilutions 
256,  512^  1024  and  2048  litres  the  molecular  conductivities 
of  malonic  acid  at  25°  are  176-8,  222-6,  269'9,  and  313-9 
respectively.  What  conclusions  may  be  drawn  from  the 
values  of  K  calculated  from  Ostwald's  dilution  law  as  to  the 
dissociation  of  the  acid  at  these  dilutions  ? 
Ans.  K  =  1-57,  1-62,  1-68,1-87  x  10  ~3,  .-.  second  dissocia- 
tion begins  to  be  appreciable  after  ca.  256  litres. 

PROBLEM  282. — A  solution  of  NaCl  containing  0-0585 
gram  per  100  c.c.  freezes  at  —  0'03674°.  The  equiv.  con- 
ductivity of  NaCl  at  infinite  dilution  is  109  r.o.  Assuming 
the  degree  of  dissociation  at  18°  to  be  the  same  as  at  0°,  cal- 
culate the  specific  conductivity  of  the  above  solution.  Take 
freezing-point  constant  =  1-89  for  gram-molecule  per  litre. 
Ans.  1-029  x  10  ~3  r.o. 

PROBLEM  283. — At  18°  the  molecular  conductivity  of  boric 
acid  at  the  dilution  v  litres  per  gram-molecule  is  //,  r.o. 
v  11-1  22-2  33-3 

I*  0-0474*  0-0670  0-0825 

Calculate  the  degree  of  dissociation  and  dissociation-con- 
stant at  each  dilution,  given  that  the  molecular  conductivity 
at  infinite  dilution  for  dissociation  into  H--  and  H2BO'3-ions 
is  346  r.o. 


ELECTROLYTIC  DISSOCIATION— PROBLEMS     141 

Ans.     a  =  1-37  x  10-*,.  1-94:  x  10  - 4,  2-39  x  10  ~4 
K  =  1-70  x  10  - 9,  1-69  x  10  - 9,  1-71  x  10  ~9. 

PEOBLEM  284. — The  specific  conductivity  of  a  0*509  N- 
KNO3  solution  at  18°  is  454  x  10  ~  *  r.o.  The  temperature- 
coefficient  of  the  specific  conductivity  between  18°  and  22°  is 
0*0208.  What  is  the  equivalent  conductivity  of  the  solution 
at  20°? 

Ans.  92-9  r.o. 

PROBLEM  285. — At  25°  the  specific  conductivity  of  malic 
acid  at  a  dilution  of  32  litres  is  1-263  x  10  ~3  r.o.  The 
equivalent  conductivity  at  infinite  dilution  for  dissociation 
into  IT-  and  C4H5O'0-kms  is  380  r.o.  Calculate  the  degree 
of  dissociation  and  the  dissociation-constant. 

Ans.   a  =  0-106,  K  =  3-93  x  10  ~4.  r- 

PROBLEM  286  (cf.  preceding  problem). — At  what  dilution     { 
the  coi 
0-02  N? 


is  the  concentration  of  H'-ions  in  a  solution  of   malic  acid 


Ans.   0-963  litre. 

PROBLEM  287.  —  At  25°  the  equivalent  conductivities  of 
fumaric  and  maleic  acids  at  a  dilution  of  32  litres  are  60'1 
and  179  r.o.  respectively.  The  equivalent  conductivities  at 
infinite  dilution  for  dissociation  into  two  ions  are  385*6  and 
391 '8  r.o.  respectively.  Compare  their  dissociation-constants. 
Aris.  Fumaric  9  x  10 -4,  maleic  1-2  x  10  -  2. 

PROBLEM  288. — The  molecular  conductivity  of  a  Ca  (NOS)2 
solution  containing  1  gram-molecule  in  23*81  litres  is  98 -9 
r.o.  at  0°.  The  molecular  conductivity  at  infinite  dilution  is 
129-2  r.o.  at  0°.  What  is  the  freezing-point  of  the  solution  ? 
Take  freezing-point  constant  =  1*89  for  gram-mole/litre. 
Ans.  -  0-2008°. 

PROBLEM  289. — The  specific  conductivity  pf  0*135  N-prop- 
ionic  acid  at  18°  is  4-79  x  10~4ro.  and  that  of  O'OOl  N- 
sodium  propionate  is  7*54  x  10  ~  5  r.o.  The  ionic  conductivity 
of  the  Na'-ion  is  44-4  and  of  the  H--ion  318  r.o.  Assuming 
the  sodium  propionate  to  be  completely  dissociated,  calculate 
the  dissociation  constant  of  propionic  acid. 
Ans.  1*41  x  10  - 5. 

*  PROBLEM  290  (cf.  preceding  problem).— The  dissociation- 
constant  of  benzoic  acid  is  6  x  10  ~5.     What  is  the  ratio  of 


142          SOLUBILITY-PRODUCT—PROBLEMS 

the  H'-ion  concentrations  in  benzoic  and  propionic  acids  at 
equal  dilutions  ? 

Ans.  2-07  :  1. 

PROBLEM  291  (cf.  preceding  problem). — At  what  dilution 
is  the  concentration  of  H'-ions  in  a  benzoic  acid  solution 
0-005  N  ? 

Ans.  2-37  litres. 

PROBLEM  292. — The  velocity-constant  for  the  inversion  of 
cane  sugar  by  0*0125  N-HC1  at  54°  is  0-00469.  The  velocity- 
constant  for  inversion  by  0*25  N-formic  acid  at  the  same 
temperature  is  0*00255.  Calculate  the  dissociation-constant 
of  formic  acid,  if  it  is  assumed  that  the  velocity-constant  is 
proportional  to  the  H'-ion  concentration,  and  that  the  HC1  is 
completely  dissociated. 

Ans.  K  =  1-9  x  10  ~4. 

PROBLEM  293  (cf.  preceding  problem). — What  would  be  the 
velocity-constant  at  54°  for  inversion  by  a  0-25  N-formic  acid 
solution,  containing  O'l  N-sodium  formate,  if  the  latter  is 
regarded  as  completely  dissociated  ? 

Ans.  0-000169. 

Solubility-Product—Hydrolysis 

PROBLEM  294. — At  18°  the  specific  conductivity  of  a 
saturated  solution  of  silver  chloride  in  water  was  2-40  x 
10  ~6  r.o.,  and  that  of  the  water  used  was  1-16  x  10  ~G  r.o. 
Given  the  equivalent  conductivity  at  infinite  dilution  of 
AgNO3  =  116-5  r.o.,  of  NaCl  =  110-3  r.o.  and  of  NaNO3  = 
105*2  r.o.,  and  on  the  assumption  that  the  AgCl  is  completely 
dissociated  in  solution,  calculate  the  solubility  in  gram- 
molecules  per  litre  and  the  solubility-product  of  AgCl  at  18°. 

Ans.  S  =  1-018  x  10-5  gram-mol./litre,  £  =  1-035  x  lO'10. 

PROBLEM  295. — At  16'3°  a  saturated  solution  of  barium 
oxalate  has  a  specific  conductivity  of  67*7  x  10  ~  6  r.o.  whilst 
that  of  the  water  used  was  1-2  x  10  ~  6  r.o.  The  ionic  con- 
ductivities for  I  Ba--  and  i  C2O"4  at  16'3°  are  50'6  and 
58-4  r.o.  respectively.  On  the  assumption  that  the  dissocia- 
tion of  the  salt  is  complete,  calculate  the  solubility,  in  gram- 
molecules  per  litre,  of  BaC,O4  at  16-3°. 

Ans.  3-05  x  10  -  * 


SOLUBILITY-PEODUCT— HYDBOLTSIS       143 

PKOBLEM  296. — At  25°  the  concentration  of  a  saturated 
solution  of  silver  acetate  is  0-0664  gram-molecule  per  litre. 
The  molecular  conductivity  of  the  solution  is  75*2  r.o.  and 
the  molecular  conductivity  of  silver  acetate  at  infinite  dilution 
is  101-5  r.o.  What  is  the  solubility-product  of  silver  acetate? 

Ans.  0-00241. 

PROBLEM  297.  —  The  solubility  of  CaSO4  at  20°  is 
2-036  grams  per  litre.  The  specific  conductivity  of  the 
saturated  solution  at  20°  is  1968  x  10  ~  6  r.o.  The  ionic 
conductivity  for  ^  Ca  •  •  at  18°  is  52  r.o.  and  the  temperature- 
coefficient  of  the  conductivity  is  0-0238.  The  ionic  con- 
ductivity for  •£  SO"4  at  18°  is  68'3  r.o.  and  the  temperature- 
coefficient  0'0227.  Calculate  the  degree  of  dissociation  of 
CaSO4  in  the  saturated  solution  and  the  solubility-product 
at  20°. 

Ans.  a  =  52-25  per  cent.,  L  =  6-11  x  10  ~5. 

PROBLEM  298. — The  solubility  of  benzoic  acid  at  25°  is 
3 '40  grams  per  litre  and  its  dissociation-constant  is  6  x  10  ~  5. 
What  would  be  its  solubility  in  a  0  01  N -sodium  benzoate 
solution  and  in  a  O'Ol  N-HC1  solution,  if  both  these  sub- 
stances are  regarded  as  completely  dissociated  ? 

Ans.  0*02678  gram-mol.  per  litre  for  both. 

PROBLEM  299. — Magnesium  carbonate,  which  is  practically 
insoluble  in  water,  dissolves  in  water  containing  C02  owing 
to  formation  of  bicarbonate.  In  a  solution  saturated  with 
CO2  under  atmospheric  pressure  the  solubility  of  MgCO3  at 
25°  was  0*325  gram-molecule  per  litre.  The  first  dissociation- 

rTT'l  TTTPO'  1 

constant  of  carbonic  acid  is  — rTT  r*r\  T       =  3-04  x  10  ~ 7  and 

[H2G(J3J 

[-TT.I  rpry  i 

the  second  dissociation-constant  is  L  rrrnrv  i*  =  1'295  x  10~n. 

L-hLUU  3J 

According  to  Henry's  law  the  concentration  of  H2CO3  is 
proportional  to  the  partial  pressure  of  COn  the  relation  at 
25°  being  given  by  [H2CO3]  =  4'92  x  10-*(OO),  if  [H2CO3] 
is  expressed  in  gram-molecules  per  litre  and  (CO2)  in  atmos- 
pheres. On  the  assumption  that  the  whole  of  the  magnesium 
in  the  solution  is  in  the  form  of  bicarbonate,  and  that  the 
bicarbonate  is  61  percent,  dissociated,  calculate  the  solubility- 
product  of  MgCO3  at  25°.  (Express  all  concentrations  in 
gram-molecules  or  gram-ions  per  litre.) 
Ans.  2-7  x  10  ~5. 


144        SOLUBILITY-PRODUCT— HYDKOLYS1S 

PROBLEM  300. — At  25°  the  molecular  conductivity  of  aniline 
hydrochloride  at  a  dilution  of  256  litres  is  130*5.  At  the 
same  dilution,  but  in  presence  of  a  sufficient  excess  of  aniline 
to  practically  prevent  hydrolysis,  it  is  107'  1.  The  conduc- 
tivity of  the  aniline  in  presence  of  its  hydrochloride  may  be 
neglected.  The  equivalent  conductivity  of  HC1  at  a  dilution 
of  256  litres  is  410.  Calculate  the  degree  of  hydrolysis  of 
aniline  hydrochloride  at  this  dilution. 

Ans.  7*72  per  cent. 

PROBLEM  301. — From  the  degree  of  hydrolysis  obtained  in 
the  preceding  problem  calculats  the  hydrolysis-constant  of 
aniline  hydrochloride,  and  the  dissociation-constant  of  aniline 
as  a  base,  given  that  the  ionic  product  for  water  at  25°  is 
1-21  x  10  ~ 14.  Assume  the  aniline  hydrochloride  and  the 
HC1  to  be  completely  dissociated,  and  the  aniline  to  be  prac- 
tically undissociated. 

Ans.  KB  =  2-52  x  10  ~5,  Kb  =  4-8  x  10  -10. 
PROBLEM  302. — From  the  result  obtained  in  the  preceding 
problem  calculate  the  degree  of  hydrolysis  of  aniline  hydro- 
chloride  in  a  O'Ol  N-solution. 

Ans.  4-88  per  cent. 

PROBLEM  303. — At  20°  the  specific  conductivity  of  a  satur- 
ated solution  of  T1C1  is  1680  x  10  ~  c  r.o.  The  total  concen- 
tration of  T1C1  as  determined  directly  at  20°  is  1*36  x  10  ~2 
gram-molecules  per  litre.  The  equivalent  conductivity  of 
T1C1  at  infinite  dilution  is  137'3  r.o.  What  is  the  degree  of 
dissociation  of  a  saturated  solution  of  T1C1  at  20°,  and  what 
is  the  solubility-product  of  T1C1  ? 

Ans.  a  =  90  per  cent.,  L  =  1-5  x  10  ~4. 
PROBLEM  304. — The  molecular  conductivity  of  carbonic  acid 
at  the  dilution  of  1  gram-molecule  in  v  litres  is  /x,  r.o. 
v  27'5  55-0  110-0 

fi    1-033  1-450  2-040. 

Calculate  the  degree  of  dissociation  and  the  dissociation- 
constant  for  each  dilution  for  dissociation  into  H--  and  HCO'3- 
ions,  given  p  for  NaHC03  =  84-9,  forNaCl  =  110-3  and  for 
HC1  =  381-9  r.o. 

Ans.    a  =  2-9  x  10  ~3,  4-07  x  10  - 3,  5*72  x  10  ~3 
K  =  3-06  x  10  - 7,  3-02  x  10  ~7,  2-99  x  10  ~7, 
mean  =  3*02  x  10  ~7. 


SOLUBILITY-PRODUCT— HYDROLYSIS        145 

*  PROBLEM  305. — From  the  dissociation-constant  of  carbonic 
acid  obtained  in  the  preceding  problem  calculate  the  hydrolysis- 
constant  for  the  hydrolysis  of  NaHCO3  according  to  the 
equation 

NaHC03  +  H,0  =  NaOH  +  H2CO3, 

and  the  degree  of  hydrolysis  of  the  salt  in  O'l  N- solution. 
Assume  the  strong  electrolytes  to  be  completely  dissociated, 
and  the  weak  acid  to  be  practically  undissociated,  and  take 
the  ionic  product  of  water  =  1*2  x  10  ~14. 

Ans.  K    =  3-97  x  10  ~  8,  x  =  0'063  per  cent. 

PROBLEM  306.— At  100°  C.  100  c.c.  of  water  dissolve  0'12 
gram  of  AgCNO.  How  much  AgCNO  will  be  dissolved  at 
this  temperature  by  100  c.c.  of  a  solution  containing  1  gram 
of  AgN03  ?  The  silver  salts  may  be  regarded  as  completely 
ionised. 

Ans.  0-016  gram. 

n   PROBLEM  307.— The  solubility  of  Ag2CO3  in  water  at  25° 
is  1  x  10  ~4  gram-molecule  per  litre.      Calculate  the  solu- 
bility-product, assuming  that  the  dissociation  is  complete. 
Ans.  [Ag-]2[CO"3]  =  4  x  10  ~12. 

PROBLEM  308  (cf.  preceding  problem). — What  is  the 
solubility  at  25°,  in  gram-molecules  per  litre,  of  Ag2CO3  in 
01  molecular  N-Na2CO3  solution.  Assume  complete  dis- 
sociation of  both  salts. 

Ans.  316  x  10  -  «. 

PROBLEM  309. — At  25°  the  solubility  of  AgCl  in  water  is 
1'5  x  10  ~ 5  gram-molecule  per  litre  and  that  of  AgBr 
7  x  10  ~  7  gram-molecule  per  litre.  On  the  assumption  that 
both  salts  are  practically  completely  dissociated,  calculate  the 
concentrations  of  Ag  •-,  01'-  and  Br'-ions  in  a  solution  which 
at  25°  is  saturated  with  both  AgCl  and  AgBr. 

Ans.  [Ag-]  =  15-03  x  10  ~  6,  [CF]  =  15  x  10 ~6, 

[Br']  =  3^26  x  10  -  8  gram-ion/litre. 

PROBLEM  310. — A  solution  of  MgCl2  was  treated  with  a 
solution  of  NH4OH  and  the  precipitated  Mg(OH)2  jfhaken 
with  the  solution  till  equilibrium  was  established1:  The 
solution  on  analysis  gave  0-0219  gram-molecule  MgCl2  per 
litre,  0-0115  gram-molecule  NH4C1  per  litre  and  0-0394  gram- 
molecule  of  free  NH4OH  per  litre.  The  dissociation-constant 
of  NH4OH  is  1-8  x  10  "  5.  Assuming  the  Mg(OH)2,  MgCl2 
10 


146  HYDEOLYSIS— PROBLEMS 

and  NH4C1  to  be  completely  dissociated  and  the  NH4OH  in 
presence  of  the  NH4C1  to  be  practically  undissociated, 
calculate  the  solubility  of  Mg(OH)2  in  water  in  gram- 
molecules  per  litre  and  its  solubility-product. 

Ans.  8  =  2-75  x  10  -  4  gram-mol./litre,  L  =  8-33  x  10  -  ". 

PROBLEM  311. — The  solubility  of  silver  hydroxide  as 
determined  directly  is  2*16  x  10  ~  4  gram-molecule  per  litre. 
The  solubility  of  AgCl  as  determined  by  the  conductivity 
method  is  1-5  x  10  ~ 5  gram-molecule  per  litre.  A  dilute 
solution  of  KOH  was  shaken  with  an  excess  of  moist  silver 
oxide  and  silver  chloride  till  equilibrium  according  to  the 
equation 

AgCl  +  KOH  =  AgOH  +  KC1 

was  established.  The  solution  then  contained  0-666  milli- 
gram-molecule of  KC1  per  litre  and  70-7  milligram-molecule 
of  KOH  per  litre.  Assuming  that  the  KC1  is  95  per  cent, 
and  the  KOH  90  per  cent,  dissociated,  calculate  the  solubility- 
product  and  degree  of  dissociation  of  AgOH  in  a  pure  aqueous 
saturated  solution. 

Ans.  L  =  2-26  x  10  ~  8,  a  =  69-6  per  cent. 
*  PROBLEM  312. — Calculate  the   percentage   hydrolysis  of 
sodium  acetate  in  0*1  N-solution  at  25°  from  the  following 
data,  assuming  that  the  salt  is  completely  dissociated. 
Dissociation-constant  of  acetic  acid  =  0-000018 
Ionic  product  for  water  =  1-21  x  10  ~  14. 

Ans.  8-2  x  10  ~  3  per  cent. 

PROBLEM  313. — At  25°  the  velocity-constant  for  the  saponi- 
fication  of  methyl  acetate  by  N-HC1  containing  c  gram-mole- 
cule of  urea  per  litre  is  k. 

c  0-0  0-5  1-0  2-0 

k  0-00315        0-00237        0-00184        0-00114. 
Assuming  that  the  velocity-constant  is  proportional  to  the 
concentration  of  free  acid,  calculate  the  hydrolysis-constant 
of  urea  hydrochloride  for  each  concentration  and  the  mean 
value. 

Ans.  0-766,  0-820,  0-772,  mean  0-786. 

p"  PROBLEM  314. — From  the  mean  hydrolysis-constant  of 
urea  hydrochloride  found  in  the  preceding  example,  calculate 
the  dissociation-constant  of  urea  into  the  ions  CON2H-5  and 
OH'.  Take  the  ionic  product  of  water  at  25°  =  T2  x  10  -  14 


HYDROLYSIS— PKOBLEMS  147 

and  assume  that  the  free  acid  and  the  salt  are  completely 
dissociated,  whilst  the  weak  base  is  practically  undissociated. 

Ans.  Kb  =  1-53  x  10  -  14. 

PROBLEM  315  (cf.  problem  162).— The  velocity-constant 
for  the  decomposition  of  diacetonealcohol  by  0-1  N-NaHS, 
under  the  same  conditions  as  in  Problem  162,  is  0-000037. 
Taking  the  degree  of  dissociation  of  01  N-NaOH  =  90  per 
cent.,  calculate  the  degree  of  hydrolysis  of  the  0-1  N-NaHS 
according  to  the  equation 

NaHS  +  H2O  =  NaOH  +  H2S. 

Assume  that  the  degree  of  dissociation  of  both  NaOH  and 
NaHS  in  the  hydrosulphide  solution  is  90  per  cent. 

Ans.  0'17  per  cent. 

*  PROBLEM  316  (cf.  preceding  problem). — What  would  be 
the  degree  of  hydrolysis  of  NaHS  in  O'Ol  N- solution  ? 

Ans.  0-537  per  cent. 

PROBLEM  317. — At  25°  the  dissociation-constant  of  aniline 
is  4-8  x  10  ~10  and  of  acetic  acid  1-8  x  10  ~  5.  The  ionic 
product  of  water  is  1-2  x  10  ~ 14.  What  is  the  degree  of 
hydrolysis  of  O'Ol  and  0*05  N-solutions  of  aniline  acetate,  if 
the  unhydrolysed  aniline  acetate  is  assumed  to  be  completely 
dissociated  ? 

Ans.  54-12  per  cent,  for  both  concentrations. 

PROBLEM  318  (cf.  preceding  problem). — What  is  the  H--ion 
concentration  in  each  of  the  aniline  acetate  solutions  ? 

Ans.  2-12  x  10  ~  5  gram-ion  per  litre  in  each. 
PROBLEM  319. — At  100°  the  following  figures  were  obtained 
in  the  catalysis  of  N/32  methyl  acetate  solution  by  N/500 
HC1,  A  being  the  titre  of  10  c.c.  of  the  solution  at  the  time  t, 

t      0         64       113      152         oo     minutes 
A  1-10     4-15     6-03     7-35     15*70  c.c.     N/50  NaOH. 
The  velocity-constant  for  the  catalysis  of  N/32  methyl  acetate 
by  a  solution  of  A1C13  containing  1/32  gram-molecule  per 
litre  was  0*00216  at  100°.     Assuming  proportionality  between 
the  velocity-constant  and  the  concentration  of  HC1,  calculate 
the  degree  of  hydrolysis  of  the  A1C13  solution. 

Ans.  8-7  per  cent. 

PROBLEM  320. — At  100°  the  following  figures  were  obtained 
in  the  catalysis  of  the  inversion  of  a  7*5  per  cent,  cane  sugar 


148  HYDEOLYSIS-  PROBLEMS 

solution- by  N/1000  HC1,  A  being  the  angle  of, rotation  at  the 
time  t, 

tQ  22  40  oo  minutes 

A   +  10-66°         4-  2-43°         -  0-49D         -  3-63°. 

At  100°  the  figures  for  the  catalysis  of  the  same  sugar 
solution  by  an  A1C13  solution  containing  1/32  molecule  per 
litre  were 

£  0  15  26  oo  minutes 

A   +  10-78°         -  0-16°         -  2-36°         -  3'53°. 

Assuming  proportionality  between  the  velocity-constant 
and  the  HC1  concentration,  calculate  the  degree  of  hydrolysis 
of  the  A1C13  solution. 

Ans.  8 '03  per  cent. 

PROBLEM  321. — The  partition-coefficient  of  the  weak  acid 
hydroxyazobenzene,  C6H5N  :  N  .  C6H4OH,  between  benzene 
and  water  is  539.  1000  c.c.  of  an  aqueous  solution  contain- 
ing O'Ol  gram-equivalent  of  the  acid  and  0-012  gram-equiva- 
lent of  Ba(OH)9  was  shaken  with  60  c.c.  of  benzene.  The 
concentration  of  the  acid  in  the  benzene  layer  was  found  to 
be  0-0537  gram  in  50  c.c.  Calculate  the  hydrolysis-constant 
of  the  barium  salt  and  the  degree  of  hydrolysis  in  a  O'Ol 
equivalent  N-solution  of  the  pure  salt. 

Ans.  KH  =  2-34  x  10  ~°,  *x  =  1-53  per  cent. 

PROBLEM  322  — At  25°  the  partition-coefficient  of  ^?-nitrani- 
line  between  benzene  and  water  is  9*0.  1000  c.c.  of  an 
aqueous  solution  containing  0 "05035  equivalent  of  HC1  and 
0-00693  equivalent  of  the  base  _p-nitraniline  was  shaken  with 
59  c.c.  of  benzene  till  equilibrium  was  established.  50 
c.c.  of  the  benzene  solution  then  contained  0-2165  gram  of 
«-nitraniline.  Calculate  the  hydrolysis-constant  of  ^-nitrani- 
line  hydrochloride  and  the  degree  of  hydrolysis  in  a  0*05035 
N  pure  aqueous  solution  of  the  salt. 

Ans.  Ka  =  10-67  x  10  - 2,  x  =  74-2  per  cent. 

PROBLEM  323. — At  25°  the  solubility  of  cinnamic  acid  in 
water  is  0'00331  gram-molecule  per  litre,  and  its  dissociation- 
constant  is  3-55  x  10  ~ 5.  The  ionic  product  of  water  is 
1-2  x  10  ~ 14.  The  solubility  of  cinnamic  acid  in  a  solution 
containing  O'Ol  gram-molecule  of  aniline  per  litre  is  0-00804 
gram-molecule  per  litre.  Assuming  the  unhydrolysed  aniline 
cinnamate  to  be  93  per  cent,  dissociated,  calculate  the  dis- 
sociation-constant of  aniline. 

A  „,      K.AO  v,  in  -  10 


GHAPTEE  X 
ELECTROMOTIVE  FORCE 

ELECTEODE  POTENTIAL.— NOEMAL  POTENTIAL.— CONCEN- 
TEATION  CELLS.— ELECTEOMOTIVE  FOECE  OF  GALVANIC 
ELEMENTS.  —  DIFFUSION  POTENTIAL.  —  OXIDATION-EE- 
DUCTION  POTENTIAL.— AFFINITY  OE  MAXIMUM  WOEK 
OF  A  EEACTION  IN  A  GALVANIC  ELEMENT.— GIBBS- 
HELMHOLTZ  EQUATION 

Electrode  Potential 

HPHE  electromotive    force  of  a   metal  electrode  (i.e.  one 
J-       which  furnishes  only  positive  ions)  against  a  solution 
of  a  salt  of  the  metal  is 

BT,      P  ET,       C 

(i)   •«•  ~^l°g^=    -SFlo&   c- 

Here  e  denotes  the  potential  difference  between  the  metal 
and  the  solution,  and  must  be  taken  with  its  proper  sign.  It 
is  positive  if  the  metal  is  charged  positively  with  respect  to 
the  solution  and  negative  if  the  metal  is  negatively  charged. 
P  is  the  electrolytic  solution  pressure  (or  tension)  of  the 
metal,  p  the  osmotic  pressure  of  the  metal  ions  in  the  solu- 
tio^  and  C  and  c  the  concentrations  of  the  metal  ions  in 
granafcions  per  litre,  corresponding  to  the  osmotic  pressures 
P  andbp.  n  is  the  valency  of  the  metal  ion  or  the  number  of 
gram-equivalents  in  a  gram-ion.  F  is  one  faraday  or  96540 
coulombs,  the  charge  carried  by  one  gram-equivalent  of  any 
ion.  T  is  the  absolute  temperature  and  E  the  gas- constant. 
When  e  is  expressed  in  volts  (i.e.  when  the  unit  of  energy  is 
the  volt-coulomb  or  joule)  the  numerical  value  of  E  is  8-32. 
On  changing  from  natural  to  common  logarithms  the  above 
expression,  therefore,  becomes 

149 


150 


NORMAL  POTENTIAL 


8-32  x  2-302  x  T,      P 
n  x  96540  g   ? 


1-984  x  lO-4  x  T,     P 


1-984  x  10-*  x  T,      G 

-  log  — 

n  &  c 

2-032  x  ET 


For  a   temperature  of  18°  C.  =  291°  absolute, 

is,  therefore,  0-0577,  and  for  25°  C.  =  298°  absolute,  it  is 
0-0591.  For  ordinary  temperatures  the  value  0'058  may  be 
employed,  so  that  we  obtain 

0-058,     P          0-058,      G 

(2)   «  =  -    log  —  = log  - . 

n        &  p  n        &  c 

For  an  electrode  which  furnishes  only  negative  ions  (e.g.  C12, 
Br2,  etc.)  the  corresponding  expression  is 

0-058,      P  0-058,      G 

(3)  «-  +  ^rlog^=  +  -irlogc- 

Normal  Potential 

The  electromotive  force,  e0,  of  an  electrode  against  a  solu- 
tion which  is  normal  with  respect  to  the  corresponding  ions 
(i.e.  contains  1  gram-ion  per  litre)  is  called  the  normal 
potential  (or  electrolytic  potential)  of  the  electrode  substance. 
Thus  for  a  metal  electrode  we  obtain  from  (i),  since  c  =^  1, 


ET 


. 
log< 


0-058 


and,  for  any  other  concentration  c,  from  (2) 

0-058 


(5) 


nF 


logo. 


E.M.F.  of  Concentration  Cell 

The  E.M.F.  of  a  concentration  cell  of  the  form 


Metal 


is 


metal  salt  solution  in 
which  the  concentra- 
tion of  the  metal 
ions  is  c 


metal  salt  solution  in 
which  the  concentra- 
tion of  the  metal 
ions  is  cl 


Metal 


E  ~  e  - 


if  the  diffusion  potential  at  the  junction  of  the  two  solutions 
is  neglected,     e  is  the  electrode  potential  at  the  junction 


ELECTEOMOTIVE  FOECE 


151 


Metal/solution  of  ionic  cone.  c. 
is  the  electrode  potential  at  the  junction 

Metal/solution  of  ionic  cono.  Cj. 
From  (i),  therefore, 


or,  at  ordinary  temceiature,  from  (2) 

,,      0-058 ,      c 
E  =  log  — 


E.M.F.  of  Galvanic  Element 

The  electromotive  force  of  a  galvanic  element  of  the  form 


Metal  M  of  j  solution  of  salt 
valency  n        of  metal  M  in 
which  the  cone, 
of    the    metal 
ions  is  c 
is,  from  (i)  and  (5), 
(7)  E  =  e  -  ei 

RT,       C 


and 


solution  of  salt 
of  metal  M.1  in 
which  the  cone, 
of  the  metal 
ions  is  cl 


RT 


Metal 
valency 


of 


. 

log.  C  -  e01  - 


are  the  potential  differences  between  the  metals 

e0  and  e01  the 


€  an    *!  are      e  poen  ncs     ew 

M  and  Mx  and  their  respective  solutions,  and 
normal  potentials  of  the  metals  M  and  Mr 


Diffusion-potential 

The  diffusion-potential  at  the  junction  of  two  solutions  of 
different  concentrations  of  the  same  binary  electrolyte,  com- 
posed- of  two  univalent  ions,  is 

1C-1A      RT 

<8>  6  =  *7T7 

c  is  the  concentration  of  the  ions  in  the  concentrated  solution 
and  Cj  that  in  the  dilute  solution.  lc  and  1A  are  the  ionic 
conductivities  of  the  positive  and  negative  ions  respectively 


152         OXIDATION-KEDUCTION   POTENTIAL 

(p.  104).  The  sign  of  e  is  that  of  the  potential  of  the  dilute 
solution  with  respect  to  the  concentrated  solution.  It  should 
be  noted  that  the  potential  of  the  dilute  solution  is  of  the 
same  sign  as  the  faster  moving  ion. 

At  the  junction  of  two  solutions  of  different  binary  electro- 
lytes of  the  same  ionic  concentration,  and  when  each  electro- 
lyte gives  only  univalent  ions,  the  diffusion-potential  is 

(9)  .- 

where  lc  and  1A  are  the  ionic  conductivities  of  the  ions  of  the 
one  electrolyte,  and  l'c  and  l'A  those  of  the  other. 

Oxidation-reduction  Potential 

Every  oxidation  process  in  which  ions  take  part  can  be  re- 
garded as  a  taking-up  of  positive  charges  or  a  giving-up  of 
negative  charges,  and,  conversely,  every  reduction  as  a 
giving-up  of  positive  or  a  taking-up  of  negative  charges.  The 
oxidation  of  ferrous  to  ferric  ions,  for  example,  may  be 
represented  by  the  equation 

Fe  •  •  +  0  =  Fe  •  •  •. 

If  such  a  process  takes  place  at  an  indifferent  (unattack- 
able)  electrode,  a  potential-difference  between  the  electrode  and 
the  solution  is  developed.  The  magnitude  of  this  potential- 
difference  depends  on  the  concentrations  of  the  substances 
taking  part  in  the  reaction.  Thus  for  the  reaction 

mA  +  nB  +  nF  =  pC  +  qD, 

in  which  m  molecules  of  A  and  n  molecules  of  B  are  converted 
into  p  molecules  of  C  and  q  molecules  of  D  by  taking  up  n 
faradays  of  positive  electricity,  the  electrode-potential  e  is 
given  by  the  equation 


The  square  brackets  denote  the  concentrations  of  the  en- 
closed substances.  e0  is  the  normal  potential  of  the  electrode 
reaction.  It  is  the  potential  of  the  electrode  against  the 
solution  when  all  the  substances  A,  B,  C  and  D,  which  de- 
termine the  potential,  are  present  in  unit  concentration.  In 
the  numerator  of  the  logarithmic  expression  are  the  concen- 
trations of  those  substances  which  are  formed  by  the  taking- 
up  of  positive  charges  (i.e.  the  higher  state  of  oxidation). 


GIBBS-  HELMHOLTZ  EQUATION  153 

Affinity  of  Reaction 

The  affinity  of  a  chemical  reaction  (p.  67)  which  takes 
place  reversibly  in  a  galvanic  element  is 

(n)  A  =  nFE  volt-coulombs  or  joules, 

where  E  is  the  electromotive  force  of  the  element  in  volts, 
F  =  96540  coulombs,  and  n  the  number  of  faradays  which 
flow  through  the  cell  during  the  transformation  of  the  quanti- 
ties of  the  reacting  substances  given  by  the  chemical  equation 
representing  the  cell-reaction.  Thus  for  the  reaction 
2H2  +  O2  =  2H20 

n  =  4,  since  4  equivalents  of  hydrogen  or  of  oxygen  are 
transformed  in  the  formation  of  2  molecules  of  water  and  1 
faraday  flows  through  the  cell  during  the  transformation  of 
each  equivalent. 

Since  1  joule  =  0*2387  calorie,  we  obtain  from  (u) 
(12)  A  =  nFE  x  0-2387 

=  nE  x  96540  x  0-2387 
=  nE  x  23040  calories. 

Gibbs-Helmholtz  Equation 

If  A  is  the  maximum  work  obtainable  from  a  chemical 
reaction  at  absolute  temperature  T  and  Q  is  the  heat  evolved 
during  the  reaction  (as  measured  in  a  calorimeter),  then  ac- 
cording to  the  Gibbs-Helmholtz  equation, 

(13)  A  =  Q  +  T^. 

-Tm  is  the  temperature-coefficient  of  the    maximum    work, 

or  the  amount  by  which  A  is  altered  for  a  temperature-change 
of  1°,   and  may   be  positive,   negative  or  zero.      A   and  Q 
should,  of  course,  be  expressed  in  the  same  unit  of  energy. 
From  (n)  and  (13)  we  obtain 


from  which  the  electromotive  force  E  of  a  galvanic  element 
at  T°  absolute  may  be  calculated  from  Q,  the  heat  of  the  re- 
action taking  place  in  the  element,  and  the  temperature- 


154       ELECTKOMOTIVE  FOEGE—  EXAMPLES 

J  777 

coefficient  of  the  electromotive  force  -^jp.     To  express  E  in 


volts,  Q  must  be  expressed  in  volt-coulombs.     If  Q  is  given 
in  calories,  then,  since  0*2387  calorie  =  1  volt-coulomb, 

Q  +  T^  Q  4-T^ 

nFx  0-2387         dT  "  n  x  96540  x  02387          dT 


E.M.F.  —  Examples 

PROBLEM  324.  —  The  normal  potential  of  Cd  is  €0  =  -  0-420 
volt  and  of  Ag  e01  =-  0*798  volt,  both  referred  to  the  N-JJ- 
electrode  as  zero.  Neglecting  the  diffusion-potential  at  the 
junction  of  the  two  electrolytes,  calculate  the  electromotive 
force  E  of  the  cell  2%--- 

Cd  |  c  =  0-5  mol.  N-Cd(N03)2  |  c,  =  0-1  N-AgNO3  |  Ag 
at  18°,  given  that  the  degree  of  dissociation  of  the  Cd(NO3)2 
solution  is  a  =  0'48  and  of  the  AgN03  solution  al  =  0-81. 

SOLUTION  324.  —  The  concentration  of  the  Cd  •  --ions  in  the 
c  =  0*5  mol.  N-solution  is  ac  =  0-48  x  0'5  gram-ion  per  litre. 
The  concentration  of  the  Ag  --ions  in  the  cl  =  0-1  N-AgN03 
solution  is  alc1  =  0-81  x  O'l  gram-ion  per  litre.  From  (7), 
therefore, 

RT. 
E  =  €°  +         lo'  ac  ~  e° 


and,  since  n  for  Cd  =  2  and  nl  for  Ag  =  1,  we  obtain  at  18° 
E  =  -  0-420  +  0-029  log  (0-48  x  0-5)  -  0*798 
-  0-058  log  (0-81  x  0-1) 

=  -  0-420  +  (0-029  x  -0-6198)  -0-798-  (0-058  x  -  1-091) 

=  -  0-420  -  0-018  -  0-798  +  0-063 

=  -  1*173  volt. 

The  Cd  is,  therefore,  the  negative  pole,  or,  in  the  cell  the 
current  flows  from  the  Cd  to  the  Ag. 

PROBLEM  325.  —  The  electromotive  force  of  the  cell 


Hg 


Hg2Cl2  in 
N-KC1 


HgOin 
N-NaOH 


Hg 


is  0-168  volt  at  18°.     The  degree  of  dissociation  of  N-KC1  and 
of  N-NaOH  is  0-72.     The  ionic  conductivity  of  K-  is  Zc=  64-9, 


ELECTKOMOTIVB  FORCE—  EXAMPLES       155 

of  OF  1A  =  65-4,  of  Na-  l'a  =  43-6  and  of  OH'  l\  =  174  at  18°. 
Taking  the  concentration  of  the  Hg  •  '-ions  in  a  saturated 
solution  of  Hg2Cl2  in  N-KG1  (normal  calomel  electrode)  as 
c  —  3  x  10  ~  20  gram-ion  per  litre,  calculate  the  solubility- 
product  of  Hg(OH)2  at  18°. 

SOLUTION  325.  —  According  to  (9)  the  diffusion-potential  at 
the  junction  of  the  two  electrolytes  KC1  and  NaOH  is 


;...-  0,58  ,og  =  0,58  log 

=  0-058  log  gjjj 

=  0-0198  volt. 

Since  OH'  is  by  far  the  fastest  moving  ion  the  KC1  solution 
will  be  negative  with  respect  to  the  NaOH  solution,  and  the 
diffusion-potential  must,  therefore,  be  subtracted  from  the 
total  electromotive  force  of  the  cell  to  obtain  the  electromotive 
force  of  the  Hg  concentration-cell  alone.  The  latter  is, 
therefore, 

E  =  0-168  -  0-0198  =  0-1482  volt.       . 

If  cl  is  the  concentration  of  the  Hg  -  --ions  in  the  NaOH 
solution  saturated  with  HgO,  we  obtain  from  (6),  since  n  =  2, 

^OJ^c  3x10-^ 

Cl  Cl 

.-.  logCi  =  2eT-684    and    ^  =  4-83  x  10  ~  2(5    gram-ion    per 
litre. 

The  solubility-product  of  Hg(OH)2  is  (p.  121) 

L  =  [Hg--][OH'f. 

In    N-NaOH    saturated   with   HgO   we    have    just    found 
[Hg  •  •]  =  Cj  =  4-83  x  10  ~  26,  and,  since  the  degree  of  dis- 
sociation of  the  N-NaOH  is  0*72,  the  concentration  of  the 
OH'-ions  is  1  x  0'72,  or  [OH'J  =  0'72.     Therefore 
L  =  (4-83  x  10  -  26)  (0-72)2 

=  2-5  x  10  -  26. 
PROBLEM  326.—  The  E.M.F.  of  the  cell 


Ag  |  c  =  0-01  N-AgN03  ci  =  °'001  N-AgNO3  |  Ag 


is  E  =••  0*0579  volt  at  25°.     Assuming  that  the  interposition 
of   saturated   NH4N03   solution   completely   eliminates    the 


156       ELECTEOMOTIVE  FOKCE— EXAMPLES 

diffusion-potential,  and  that  the  0*001  N-AgN03  solution  is 
completely  dissociated,  calculate  the  degree  of  dissociation 
and  the  concentration  of  the  Ag  --ions  in  the  0*01  N-AgN03 
solution. 

SOLUTION  326. — If  a  is  the  degree  of  dissociation  of  the 
c  =  0-01  N-solution,  the  concentration  of  the  Ag  --ions  is  ac. 
Since-  the  cl  =  0*001  N-solution  is  completely  dissociated, 
the  concentration  of  the  Ag  --ions  in  it  is  cr  For  the  E.M.F. 
of  the  Ag  concentration-cell  we  therefore  obtain  from  (6) 

ET, 


and,  since  n  =  1,  at  25C 


0-0579  =  0-059  log  - 


x  0-01 


0-001 

/.    a  =  0-958. 

The  concentration  of  the  Ag  --ions  in  the  c  =  O'Ol  N-solu- 
tion is 

[Ag  ']  =  ac  —  0*958  x  O'Ol  =  0*00958  gram-ion  per  litre. 

PROBLEM  327. — The  normal  potential  of  silver  is  eAg  = 
+  0*771  volt  (hydrogen  standard),  that  of  chlorine  at  atmo- 
spheric pressure  is  €C12  =  +  1-366  volt,  and  that  of  bromine  at 
atmospheric  pressure  is  eBr2  =  +  0'99  volt.  The  solubility- 
product  of  AgCl  at  25°  is  L^  =  2  x  10  ~  10,  and  that  of  AgBr 
L2  =  2  x  10  ~13.  What  is  the  affinity  of  silver,  in  joules  and 
calories,  (a)  to  chlorine  under  atmospheric  pressure,  (b)  to 
bromine-vapour  under  atmospheric  pressure  ? 

SOLUTION  327. — (a)  The  affinity  of  silver  to  chlorine  at 
atmospheric  pressure  is  denned  as  the  maximum  work  which 
can  be  gained  by  the  reversible  combination  of  1  molecule  of 
chlorine  at  atmospheric  pressure  with  2  equivalents  of 
metallic  silver  to  form  two  molecules  of  solid  silver  chloride, 
according  to  the  equation  C12  +  2Ag  =  2AgCl. 

This  reaction  proceeds  reversibly,  and  therefore  yields  the 
maximum  work,  in  a  galvanic  element  of  the  type 


Ag 


saturated  solution  of  AgCl 


Pt  charged  with  C12  at 
atmospheric  pressure. 


When  the  element  is  in  action,  Ag  goes  into  solution  as 
Ag'-ions  at  one  electrode  and  C18  as  Cl'-ions  at  the  other,  and 


ELECTROMOTIVE  FORCE—  EXAMPLES        157 

solid  AgCl  is  formed.  If  E  denotes  the  E.M.F.  of  fcho  ele- 
ment, then  by  the  solution  of  1  molecule  of  C12  and  2  equi- 
valents of  Ag  the  electrical  energy 

A  =  2#  x  96540  joules 

is  produced.  E  can  be  calculated  from  the  two  separate 
electrode  potentials,  ex  between  the  silver  electrode  and  the 
solution  and  e2  between  the  chlorine  electrode  and  the  solu- 
tion. 

From  (i)  and  (3) 

RT,       [Ag-] 

«  -  +      *°          and  •  -' 


CAg  and  CC12  are  the  concentrations  of  the  Ag  •-  and  Cl'-ions 
corresponding  to  the  electrolytic  solution  pressures  of  silver 
and  of  chlorine  at  atmospheric  pressure.  The  square 
brackets,  as  usual,  denote  the  concentrations  of  the  ions  at 
the  electrodes.  Since  in  both  cases  n  =  1,  we  obtain  from  (7) 

ET        [Ag-]      ET         CC12 


ET  ET  ET 

=  -jr  log,  [Ag  •]  [Cl'J   -  -p  log.  0M-  -jr  log,  CC12 


eAg-  €C12. 

The  product  [Ag  •]  [01']  is  equal  to  the  solubility-product 
L1  of  silver  chloride,  since  the  solution  is  saturated  with 
AgCl.  Therefore 

ET 
E  =  -p  loge  L1  +  e^  -  €C12 

and,  at  25°, 

E  =  0-059  log  2  x  10-10  +  0-771  -  1-366 
=  0-059  x  -  9-699  -  0-595  =  -  1-167  volt. 

The  negative  sign  means  that  the  silver  electrode,  which 
in  the  calculation  was  taken  as  positive  (  +  ej,  is  the  negative 
pole.  In  the  calculation  of  A  the  sign  of  E  is  of  no  conse- 
quence ;  it  is  only  the  absolute  value  that  is  required. 

The  affinity  of  silver  to  chlorine  at  atmospheric  pressure 
is,  therefore, 


158       ELECTKOMOTIVE  FORCE— EXAMPLES 

A  =  2  x  1-167  x  96540  =  225300  joules 
or,  in  heat  units, 

A  =  2  x  1-167  x  23040  =  53760  calories. 

(b)  For  the  affinity  of  silver  to  bromine-vapour  at  atmo- 
spheric pressure  we  have  the  corresponding  equation 

A  =  ZE  x  96540  joules, 
where  E  is  the  E.M.F.  of  the  galvanic  element 


Ag 


solution  saturated  with 


AgBr 


Pt  charged  with  Br- vapour  at 
atmospheric  pressure. 


As  above 


7?  *~.1  T 

^    =   ~  10g*  L2    +    €Ag    ~    *Br2 

=  0-059  log  2  x  10  -  13  +  0-771  -  0-99 
=  0-969  volt, 
and 

A  =  2  x  0-969  x  96540  =  187100  joules. 
=  2  x  0-969  x  23040  =  44640  calories. 

PBOBLEM  328. — The  solubility  of  silver  chloride  in  water 
at  25°  0.  is  S  =  1-4  x  10  ~  5  gram-molecules  per  litre.  The 
solubility  in  ammonia  solutions  is  greater,  and  is  proportional 
to  the  total  concentration  of  the  ammonia.  The  factor  of 
proportionality  is  k  =  0*05.  The  increased  solubility  in 
ammonia  depends  on  the  formation  of  a  complex  ion  accord- 
ing to  the  scheme  Ag  •  -f  nNH3  =  Ag(NH3)n-.  How  many 
molecules  of  ammonia  take  part  in  the  formation  of  this  com- 
plex cation,  i.e.  what  is  the  value  of  n,  and  what  is  thet dis- 
sociation-constant K  =  nf  AJJJ  \3.-|  °f  tne  complex  ion  ? 

SOLUTION  328. — In  any  solution  which  is  saturated  with 
silver  chloride  the  product  of  the  concentrations  of  the  ions, 
[Ag*]  [OF],  must,  for  a  given  temperature,  have  a  definite 
constant  value  L,  which  is  called  the  solubility-product  of 
silver  chloride.  In  a  saturated  aqueous  solution  which  con- 
tains only  silver  chloride,  [Ag  •]  =  [01']  =  S,  and,  therefore, 
L  =  S2,  it  we  assume  that  the  dissociation  of  the  dissolved 
salt  is  complete.  The  greater  solubility  of  silver  chloride  in 
solutions  of  ammonia  compared  with  pure  water  is  caused  by 


ELECTROMOTIVE  FORCE—  EXAMPLES        159 

the  formation  of  a  complex  cation  by  the  addition  of  ammonia  - 
to  the  Ag  '-ions  according  to  the  scheme 

Ag-  +  wNH3  =  Ag(NH3)n-; 

this  diminishes  the  concentration  of  the  silver  ions,  and, 
therefore,  requires  an  increase  in  the  concentration  of  the 
chlorine  ions,  so  that  the  value  of  the  solubility-product  may 
be  maintained.  The  application  of  the  law  of  mass-action  to 
the  formation  of  the  complex  cation  leads  to  the  equilibrium- 
equation 

[Ag-][NH3]-  _ 

w  [Ag(NH3)n-]  = 
n  must,  of  course,  be  a  whole  number. 

The  total  concentration  of  silver  in  the  solution,  a,  is  equal 
to  the  sum  of  the  concentrations  of  the  Ag'-ions  and  the 
Ag(NH3)n  •  -ions,  if  we  assume  that  the  salts  AgCl  and  the 
Ag(NH3)nCl  are  practically  completely  dissociated.  Accord- 
ing to  the  conditions  of  the  problem,  the  solubility  of  AgCl  in 
the  ammonia  solution  is 

(2)  a  =  [Ag  •]  +  [Ag(NH8)n-]  =  fc(NH3). 

Here  (NH3)  denotes  the  total  concentration  of  ammonia  in 
the  solution,  and,  therefore, 

(3)  (NH3)  _  [NHJ  +  [NH4-]  +  «  [Ag(NH3)n-], 

since  the  total  ammonia  in  the  solution  is  equal  to  the  sum 
of  the  undissociated  ammonia,  the  ammonium  ions  and  the 
ammonia  in  the  complex  ion.  From  (2)  we  therefore  obtain 

(4)  [Ag-]  +  [Ag(NH3)n-]  = 


When  the  concentration  of  ammonia  is  sufficiently  great,  the 
concentration  of  the  silver  ions,  [Ag*],  which  is  always  less 
than  £,  that  is,  less  than  1'25  x  10  ~  5,  may  be  neglected  in 
comparison  with  the  total  concentration  of  silver  a.  Under 
these  circumstances  we  obtain  from  (2) 

a=  [Ag(NH3)M-], 
and  hence  from  (4) 

a  =  &[NH3]  +  &[NH4  •]  +  Jena, 
a(l  -  kn)  =.  M[NH8]  +  [NH4  •]} 
and 


160       ELECTROMOTIVE  FORCE— EXAMPLES 
By  dividing  the  equation 

[Ag-][cr]  =  i 

by  equation  (1)  written  in  the  form 
[Ag.][NH3]*  _ 

a  ' 

we  obtain 

L 
"K' 

Every  molecule  of  AgCl  which  dissolves  produces  one 
Ag(NH3)n  --ion  and  one  Cl'-ion.  The  concentrations  of  these 
ions  are,  therefore,  equal  and  [01']  =  a,  and  (6)  becomes 

w  -  § 

and,  therefore, 

(7)  a- 

When  equations  (5)  and  (7)  are  compared,  it  is  evident, 

since  -, 7—  and  A-^  are  constants,  that  for  different  con- 

1  —  Kn  \  K 

centrations  of  ammonia  the  two  equations  can  be  simultane- 
ously satisfied  only  (1)  if  the  concentration  of  the  ammonium 
ions,  [NH4  •],  is  negligible  compared  with  the  concentration 
of  the  undissociated  ammonia  molecules,  [NHJ,  a  condition 
which  is  satisfied  in  the  case  of  weak  bases  like  ammonia, 
and  (2)  if  n  =  2.  From  (5)  and  (7)  we  then  obtain 


and 

2fr2       1-42  x  10  -  10  x 


k*  (O05)2 

=  5-4  x  10  ~8. 

The  silver-ammonia  complex  ion  has,  therefore,  the  com- 
position Ag(NH3)2  •,  and  its  dissociation  -constant  at  25°  C.  is 
54  x  10  ~8. 


ELECTROMOTIVE  FORCE—  EXAMPLES        161 

PKOBLEM   329   (cf.  preceding  problem).  —  What    are    the 
E.M.F.'s,  El  and  E2  of  the  cells 

Ag/c  N-AgNO3/c'  N-KC1,  AgCl/Ag 
and 

c  N-AgNO 


C  N-NH 


c'  N-KC1,  AgCl 


Ag 


if  o  =  0-01,  c'  =  1  and  C  =  1  ? 

SOLUTION  329. — According  to  Nernst's  formula  (equation 
(6))  the  E.M.F.  of  a  concentration  cell  is 


nF     ='  [Me-]2' 

if  the  diffusion-potential  at  the  point  of  contact  of  the  two 
electrolytes,  in  which  the  electrodes  are  placed,  is  neglected. 
n  is  the  valency  of  the  cation,  [Me*^  and  [Me  ]2  the  concentra- 
tions of  the  cation  at  the  two  electrodes.  The  cells  described 
in  the  problem  are  to  be  regarded  as  concentration  cells  with 
respect  to  the  Ag  •  cation,  and,  since  n  =  1  in  this  case,  the 
formula  becomes 


The  concentrations  of  the  silver  ions,  [Ag  •],  in  the  various 
solutions  may  be  calculated  as  follows. 

In  the  c  N-AgNO3  solution  the  silver  ion  concentration, 
[Ag  'Jj,  is  approximately  equal  to  the  total  concentration, 
since  the  salt  may  be  regarded  as  practically  completely 
dissociated, 

.-.[Ag  •],  =  <;. 

In  c'  N-KC1  solution,  which  is  saturated  with  AgCl,  the 
product  of  the  silver  ion  concentration,  [Ag  -].2,  and  the 
chlorine  ion  concentration,  [OF],  must  be  equal  to  L,  the 
solubility-product  of  AgCl  (cf.  problem  328).  If  the  KC1  is 
regarded  as  completely  dissociated,  [Cl'J  =  c'.  We  obtain, 
therefore, 

[Ag  -]2  rCl']  =  L, 


and,  for  the  E.M.F.  of  the  first  cell  at  25°  C., 

BT,       cc' 


n 


162       ELECTROMOTIVE  FORCE—  EXAMPLES 

P  -  0-069  log  ax°g.10-  0-454  volt. 

The  concentration  of  the  silver  ions,  [Ag  •],  in  the  C 
N  -ammonia  solution  follows  from  the  equation  (cf.  problem 
328) 

a)[Ag-J[NHJ«_g 

[Ag(NH3)2-J 

Since  K  has  a  relatively  small  value  (5*4  x  10  ~  8),  the 
complex-formation  between  the  silver  ions  and  the  excess  of 
ammonia  may  be  regarded  as  practically  complete,  i.e. 
practically  the  whole  of  the  silver  in  the  solution  is  present 
as  the  complex  ion  Ag(NH3)2', 

.-.  [Ag(NH3)2-]  =  c. 
The  concentration  of  the  free  ammonia  is,  therefore, 

[NHJ  =  C-2o, 
and,  from  equation  (1),  the  concentration  of  the  silver  ions  is 


For  .Eg,  the  E.M.F.  of  the  second  cell,  we  therefore  obtain 


=  0*0264  volt. 

The  addition  of  C  =  1  N-ammonia  to  the  0-01  N-AgNO3 
solution  has,  therefore,  diminished  the  E.M.F.  by  0'454  - 
0-0264  -  0-428  volt.  The  sign  of  the  E.M.F.,  however,  re- 
mains unchanged. 

PROBLEM  330. — The  E.M.F.  of  the  hydrogen- oxygen  cell 
at  25°  is  El  =  1-23  volt,  and  the  E.M.F.  of  the  cell 

Ag/saturated  solution  of  Ag2O/H2 

is  E<2  =  1-18  volt.  What  is  the  affinity  of  silver  to  oxygen 
under  atmospheric  pressure  (a)  at  25°  and  (b)  at  35°,  if  the 
molecular  heat  of  formation  of  Ag2O  is  Q  =  6400  cals.  ? 

SOLUTION  330.-  -(a)  The  E.M.F.  E  of  any  galvanic  element 


ELECTROMOTIVE  FORCE—  EXAMPLES       163 

which  works  reversibly  is  a  measure  of  the  affinity  A  of  the 
reaction  which  produces  the  current  ;  for 

A  =  nE  x  96540  volt-coulombs, 

if  n  denotes  the  number  of  equivalents  of  ions  which  enter,  or 
are  deposited  from,  the  solution  during  the  transformation  of 
the  quantities  of  the  reacting  substances  given  by  the  equation 
representing  the  cell-reaction.  The  E.M.R  of  the  hydrogen- 
oxygen  cell 

H2/any  aqueous  solution/O2 
is,  therefore,  a  measure  of  the  affinity  of  the  reaction 

(1)  2H2  +  02  =  2H20, 
and  the  E.M.F.  of  the  cell 

H2/saturated  solution  of  Ag2O/Ag 
is  a  measure  of  the  affinity  of  the  reaction 

(2)  H2  +  Ag2O  =  H2O  +  2Ag. 
By  multiplying  (2)  by  2  and  subtracting  from  (1)  we  obtain 

(3)  02  +  4Ag  =  2Ag20, 

or,  for  the  affinities, 

A3  =  AI  —  *A<I. 
Now 

Al  =  4^  x  96540  volt-coulombs, 

and  AZ  =  2.EJ2  x  96540  volt-coulombs, 
/.  As  =  ±(El  -  E2)  96540  volt-coulombs, 
=  0-20  x  96540  volt-coulombs, 
=  0-20  x  23040  =  4608  calories  at  25°  C. 

(6)  To  calculate  the  affinity  at  35°  C.  we  use  Helmholtz's 
equation  (13) 

mdA        dA       A  -  Q 

A  -  Q  +  Tdr  or  dr  -  -r*1 

Q,  the  heat  of  formation  of  2  molecules  of  Ag2O,  is  12800 
calories,  therefore 


By  raising  the  temperature  1°  C.  the  affinity  is  diminished 
by  27-5  cals.,  and,  therefore,  for  a  rise  of  10°  C.  by  275  cals., 


164       ELECTROMOTIVE  FORCE—  EXAMPLES 

if  the  temperature-coefficient  is  assumed  to  be  constant  in  the 
small  interval  between  25°  and  35°.  The  affinity  of  silver  to 
oxygen  at  35°  is,  therefore, 

4608  -  275  =  4333  cals. 

PROBLEM  331.  —  The  potential  of  a  mercury  electrode  in 
a  c  =  1  N-KC1  solution  which  is  saturated  with  calomel  is 
€l  =  +  O283  volt,  and  the  solubility-product  of  calomel, 
Hg2Cl2,  is  L  =  [Hga--]  [Cl']2  =  3-5  x  10  ~  18.  When  a 
c'  =  0-001  mol.  N-FeCl3  solution  is  shaken  with  solid  calomel 
and  metallic  mercury,  reduction  of  the  ferric  chloride  pro- 
ceeds until  the  ratio  of  ferric  to  ferrous  ions  in  the  solution 
becomes  [Fe  •  •  ']/[Fe  •  •]  =  x.  What  is  the  value  of  a?,  if  the 
normal  oxidation-potential  of  a  ferric-ferrous  ion  electrode  is 
€o  =  0-743  volt? 

SOLUTION  331.  —  Let  the  ratio  of  ferric  to  ferrous  ions  in 
the  solution  at  equilibrium  be  x.  Then  the  oxidation-potential 
of  the  solution  is,  from  (10), 

BT, 

€  =  €0  +   -p-  log.  X  VOlt, 

since  n  =  1. 

The  mercury  potential  in  the  solution  at  equilibrium  must 
have  the  same  value,  therefore 

(1)   €0 

where  e'0  denotes  the  normal  potential  of  mercury  against 
mercurous  ions,  and  y  the  concentration  of  the  divalent 
(n  =  2)  mercurous  ions,  [Hg2  •  •],  in  the  solution  at  equili- 
brium. 

In  equation  (1)  e'0,  x  and  y  are  unknown,  and  therefore 
two  further  equations  are  necessary  to  enable  us  to  calculate 
them.  From  the  potential  of  the  normal  calomel  electrode 
€j  we  can  obtain  €0  since 

7?  T*  7?  T*  T 

(2)   €l  =  e'0  + 


,  ET 
h  2F 

if   the   potassium    chloride   is   regarded    as    completely   dis- 
sociated. 


ELECTROMOTIVE  FORCE—  EXAMPLES       165 

y  may  be  calculated  from  the  extent  to  which  the  ferric 
chloride  solution  is  reduced.  The  reduction  takes  place  ac- 
cording to  the  equation 

2  Fe  •  •  •  +  2Hg  =  2Fe  •  •  +  Hg2  •  •, 
or  2FeCl3  +  2Hg  =  2FeCl2  +  Hg2Cl2. 

For  every  2  molecules  of  FeCl3  reduced,  1  molecule  of 
Hg2Cl2  is  formed,  and,  since  Hg2012  is  practically  insoluble, 
it  is  deposited,  and  2Cl'-ions  disappear  from  the  solution. 

Now  [Fe  •  •  •]  +  [Fe  •  •]  =  c',  the  total  concentration  of  iron 
in  the  solution, 

[Fe  •  •  •]  c' 

and  r-Tn   1  11  =  x,  therefore  [Fe  •  •]  =  *  —  -  —  . 
L  jj  e  *  *j  i  +  x 

In  the  original  ferric  chloride  solution  there  were  3d'-ions 
for  every  Fe  •  •  '-ion  ;  the  concentration  of  the  Cl'-ions  was, 
therefore,  3c.  In  the  solution  at  equilibrium,  for  every  Fe  •  •  •- 
ion  which  has  been  reduced,  or  for  every  Fe  •  *-ion  which  has 
been  formed,  ICl'-ion  has  disappeared  from  the  solution.  In 
the  solution  at  equilibrium  there  remain,  therefore, 

o  ,  c'          2c/  +  Sxc'       . 

3c   -  1  —  -  —  =  —  1—  -      -  Cl  -ions. 
1  +  x          1  +  x 

Since  the  solution  is  saturated  with  Hg2Cl2,  the  equation 

[Hg2--][Cl'p  =  £ 
must  hold,  and,  therefore, 


(3) 


(C') 


If  in  equation  (1)  we  substitute  the  value  of  e'0  obtained  in 
equation  (2)  and  the  value  of  y  obtained  in  equation  (3),  we 
obtain 

ET,  ET,       L      ET,          L(l  +  xf 

.  +  T  loge  x--€l-  wloge  -  +  ^loge  (V)2V(2  +  3;)2> 

or,  collecting  the  logarithmic  terms  on  one  side  and  dividing 
throughout  by  RTjF, 

»JL(l  +  x) 


inc 
°°ec'(2  +  3x)xjL       BT/F' 


166      ELECTEOMOTIVB  FOECE—  EXAMPLES 


e  =  2-718  is  the  base  of  the  natural  logarithms.  The 
right-hand  side  of  equation  (4)  contains  only  known  quanti- 
ties. If,  for  brevity,  we  call  its  value  At  we  obtain  from  (4) 

1  +  x  =  #(2  +  3x)A  =  2Ax  +  3Ax2, 


6A       '   \V~  64     ;    '   34' 
The  value  of  A  is  obtained  from  the  equation 

eO-ei  />'  0-748-02^3  10-3 

A  =  e  -WF  x  -  =  (2-718)     0.059      x  J±_ 
c  1 

=  (2-718)18  x  ID-3 
=  6-56  x  104. 

Since  A  is  very  large,  x  must,  according  to  equation  (4)» 
be  very  small.  If  in  this  equation  we  neglect  the  term  in  x2 
as  small  compared  with  x,  we  obtain  approximately 


The  reduction  of  the  ferric  chloride  to  ferrous  chloride  is, 
therefore,  practically  complete. 

PROBLEM  332. — With  potassium  iodide  mercuric  ions  form 
the  complex  ion  HgI4".      The  dissociation-constant  of   the 

complex  ion  is  K  =  fe — °^  j  L  =  5  x  10~81.  In  presence 
of  metallic  mercury  equilibrium  between  mercurous  and 

mercuric  ions  is  reached  when  I    &2   j  =  JT,  =  120.    (1)  How 

[Hg '  J 

many  grams  of  mercurous  iodide  dissolve  on  shaking  solid 
mercurous  iodide  with  a  litre  of  c  =  1  N-potassium  iodide 
solution,  if  the  solubility  of  Hg2I2  is  a  =  3'1  x  10~10  gram- 
molecule  per  litre?  (2)  What  is  the  concentration  of  a  KI 


ELECTROMOTIVE  FORCE— EXAMPLES       167 

solution  which  dissolves  equal  weights  (in  grams)  of  Hg  in 
the  form  of  mercurous  and  mercuric  salts  ? 

SOLUTION  332. — (1)  In  any  solution   which   is   saturated 
with  mercurous  iodide  the  equation 

[Hg2  •  •]  [17  =  L 

must  be  satisfied,  where  L  is  the  solubility-product  of  Hg2I2. 
In  a  pure  saturated  aqueous  solution  of  Hg2I.;,  in  which  the 
salt  may  be  regarded  as  completely  dissociated, 

[Hg2-  •]  =  a,  and  [I']  =  2a, 

(1)  .-.  [Hg2.-][IT  =  4a3  =  £. 

Mercurous  ions  react  in  presence  of  metallic  mercury  accord- 
ing to  the  equation  Hg2*  •  =  Hg  +  Hg  •  •,  until  equilibrium  is 
reached  when 

-_(2)gtq  =  *, 

If  the  solution  contains  potassium  iodide,  the  potassium  iodide 
reacts  with  the  mercuric  ions  to  form  the  complex  ion  HgI4", 
according  to  the  equation 


The  equilibrium  equation  for  this  reaction  is,  therefore, 


In  the  case  of  the  potassium  iodide  solution  which  is  satu- 
rated with  mercurous  iodid0,  equations  (1),  (2)  and  (3)  must 
be  simultaneously  satisfied.  In  these  equations  [Hg2«  •], 
[Hg  •  •],  [HgI4"J  and  [I']  are  unknown.  An  additional  equa- 
tion is,  therefore,  required  for  their  solution.  This  is  obtained 
as  follows  from  the  initial  concentration  of  KI,  which  is 
known.  If  the  concentration  of  the  undissociated  molecules 
is  neglected,  i.e.  if  all  the  salts  are  regarded  as  completely 
dissociated,  the  total  iodine  in  the  solution  is  equal  to  the 
sum  of  the  iodine  present  as  I'-  and  HgI4"-ions.  Of  the  four 
iodine  atoms  contained  in  HgI4",  only  two  are  derived  from 
the  dissolved  Hg2I2,  and  the  other  two  from  the  KI  originally 
present.  The  concentration  of  the  free  iodine  ions  is,  there- 
fore, 

(4)  [I']  =  c  -  2[HgI4"]. 
By  dividing  (1)  by  (2)  we  obtain 


168         ELECTROMOTIVE  FORCE—  EXAMPLES 

(5)  [Hg  •  •]  [IT  =  |, 
and  from  (3)  and  (5) 

(6)  ^[rp- 

From  (6)  and  (4)  we  obtain 
(7) 


(7)  is  a  quadratic   equation  containing   only   one  unknown 
quantity  [I'J,  which  may,  therefore,  be  calculated  : 


,8)   rri  _        KK, 

+ 
According  to  (7) 


If  x  grams  of   mercurous  iodide   dissolve,  and   if  M  is  the 

2;£ 
molecular  weight  of  Hg2I2,  there  are  formed  —  equivalents  of 

HgI4",  if  the  concentrations  of  the  Hg2-  •-  and  Hg  •  •  -ions  in 
the  concentrated  potassium  iodide  solution  are  neglected. 
From  (7)  we  therefore  obtain 


(9)  x  =  ^     ^l^  grams. 
Substituting  the  numerical  values  in  (8)  we  obtain 

rri  =     -  5  x  io~ai  x  120 

4  x  4  x  (3-1  x  10 " 10)3 

(5  x  10  ~ 31  x  12Q)2  5  x  10- 31  x  120  x  1 

16  x  16  x  (3-1  x  10 -10)6      2  x  4  x  (3'1  x  10  ~lof 
=   -  0-126  +  V(0-126)2  +  U 
=  0-392, 
and  from  (9) 

=  554(1  -  0-392)  = 
4 

(2)  The   ratio   of   mercurous  to  mercuric   ions  is  always 
K-i=  120.     [Hg-  •],  may,  therefore,  be  neglected  in  compari- 


ELECTEOMOTIVE  FORCE—  EXAMPLES       169 

son  with  [Hg2*  •],  and  the  whole  of  the  mercuric  mercury  may 
be  regarded  as  contained  in  the  complex  ion.  If,  therefore, 
equal  weights  of  mercury  dissolve  in  the  mercuric  and  mer- 
curous  forms,  the  weight  of  mercury  in  the  mercuric  complex 
HgI4"  must  be  equal  to  that  in  the  mercurous  ions,  or 

(10)  [HgI4"]  = 
From  (3)  and  (10) 

[Hg  ••][!']*  = 

and  from  this  equation  and  (2) 
[I']'  = 


From  (10)  and  (1)  we  obtain 


=  2-18  x  10  ~14, 
and  from  (4) 

c  =  [I']  +  2[HgI4"]  =  740  x  10  ~s  +  2-18  x  10  ~14 
=  7*40  x  io~8N. 

In  all  potassium  iodide  solutions  of  measurable  concentration, 
therefore,  the  mercurous  iodide  dissolves  chiefly  as  complex 
mercuric  salt. 

PROBLEM  333.  —  The  normal  potential  of  a  copper  electrode 
against  cupric  ions  is  e0  =  +  0*329  volt,  the  constant  for  the 
equilibrium  between  cupric  and  cuprous  ions  in  presence  of 
metallic  copper  is  [Cu  -]'2/[Cu  •  •]  =  K  =  0-6  x  10  ~  3  and  the 
solubilities  of  cuprous  chloride  and  bromide  are  a  =  1-1  x 
10  "  3  and  b  =  2*04  x  10  ~4  gram-molecules  per  litre  respec- 
tively. What  is  the  potential  of  a  copper  electrode  (1)  in  a 
c  =  0-1  N  -  HC1  solution  which  is  saturated  with  cuprous 
chloride,  and  (2)  in  a  c  =  0-1  N  -  HBr  solution  which  is 
saturated  with  cuprous  bromide  ?  (3)  What  is  the  normal 
potential  e'0  of  a  copper  electrode  against  cuprous  ions  ? 

SOLUTION  333.  —  (1)  and  (2).  In  the  solution  which  is  sat- 
urated with  CuCl,  and  which  at  the  same  time  is  in  equi- 
librium with  metallic  copper,  let  the  concentrations  of  the 
cupric  (Cu  •  •)  and  cuprous  (Cu  •)  ions  be  x  and  x  respectively, 
and  in  the  corresponding  CuBr  solution  let  the  concentrations 


170       ELECTKOMOTIVE  FOECE— EXAMPLES 

of  the  cupric  and  cuprous  ions  be  y  and  y  respectively.  Then 
the  potential  of  a  copper  electrode,  regarded  as  a  cupric 
electrode,  against  the  CuCl  solution  is 

(1)  £l  =  c0  +  ~logex, 

and  against  the  CuBr  solution 

•pm 
(2)c2  =  <o  +  ~ 

x  and  y  may  be  determined  from  the  following  considera- 
tions : — 

1.  The  cupric  and  cuprous  ions  are  in  equilibrium  with  one 
another  and  with  metallic  copper  according  to  the  equation 

Cu • • +  Cu  =  2Cu  -, 
therefore, 

(3)  (x'Y  =  Ex 
and 

(4)  (yj  =  Ky. 

2.  The  solutions  are  saturated  with  CuCl  and  CuBr  re- 
spectively, therefore  [Cu  •]  [Cl'J  =  Ll  and  [Cu  •]  [Br']  =  £2, 
where  Ll  and   L.2  are  the   solubility-products  of  CuCl  and 
CuBr.     Ll  and  L2  are  equal  to  the  squares  of  the  solubilities 
of  CuCl  and  CuBr  in  water,  if  the   salts  are  regarded  as 
completely  dissociated.      According  to  the   law   of   electro- 
neutrality 

[Cl']  =  c  +  [Cu  •]  +  2[Cu  •  •]  =  c  +  x  +  %x 
and 

[Br]  =  c  +  [Cu  •]  +  2[Cu  •  •]  =  c  +  y'  +  2y, 
therefore, 

(5)  x'(c  +  x'  +  2a)  =  a2  =  Llt 
and 

(6)  y'(c  +  y'  +  2y)  =  6*  =  Lr 

Equations  (3)  to  (6)  are  sufficient  for  the  evaluation  of  x, 
y,  x'  and  y'. 

From  (3)  x  =  &  and  from  (4)  y  =  (^J2.       Putting   these 
K  K 

values  of  x  and  y  into  equations  (5)  and  (6)  we  obtain 
(7)  x(c  +  x  +  2< 


ELECTEOMOTIVB  FOECE—  EXAMPLES       171 

and 

(8)  y'c  +  y' 


Since,  according  to  the  conditions  of  the  problem,  x  is  less 

<2(x'\'2 

than  a  and  a  is  very  small,  x'  +    ^    '    may  be  neglected    in 

comparison  with  c.     Similarly  y'  +  -^P-  may  De  neglected 

in  comparison  with  c. 

From  (7)  and  (8)  we  therefore  obtain 

a2  62 

x'  =  —  and  y  =  -  , 
c  c 

and  hence  from  (3)  and  (4) 


Putting  these  values  of  x  and  ?/  into  equations  (1)  and  (2)  we 
obtain 

,   ET 


and 

64  .   ET 


and,  substituting  the  numerical  values, 

e,  =  0-329  +  0-059  log    C1'1  x  1Q  ^T 
80-lVo-6  x  10  - 


=  0-329  +  0059  x    -  3'31 
=  0*134  volt. 

=  0-329  +0-059  log 


-^0-6  x  10  -3 
=  0-329  +  0-059  x   -  4-77 
=  0*047  volt. 

(3.)  The  normal  potential,  e'0,  of  copper  against  a  solution 
of  cuprous  ions  is  obtained  as  follows  :  the  potential  of  a 
copper  electrode  against  an  x'  N-solution  of  cuprous  ions  is 


172       ELECTEOMOTIVE  FORCE— EXAMPLES 

In  the  solution  saturated  with  cuprous  chloride  ej  =  +  0-134 


volt  and  x'  =  —  ,  therefore, 

c 


TtT 
i  -  ~p  log«  x' 

RT 


0-134  -  0-059  log 


=  0-134  +  0-059  x  4-92 
=  0*424  volt. 

PROBLEM  334. — The  potential  between  the  terminals  of  a 
lead  accumulator,  which  is  filled  with  100  c.c.  of  molecular 
N-H2SO4,  is  E  =  1-88  volt  at  25°.  The  normal  potential 
of  lead  is  €0  =  —  0-12  volt,  the  normal  oxidation-potential 
of  a  plumbic-plumbous  ion  electrode  is  e'0  =  +  1-80  volt, 
and  the  solubility  of  lead  sulphate  is  a  =  1-26  x  10  ~4  gram- 
molecule  per  litre.  What  is  the  concentration  of  plumbic 
ions  at  the  lead  peroxide  electrode  (1)  when  the  accumu- 
lator is  on  open  circuit,  (£)  after  a  discharge  of  1  ampere- 
hour  ?  (3}  What  is  the  potential  between  the  terminals  of 
the  accumulator  after  this  discharge  ? 

SOLUTION  334. — (1)  The  chemical  process  which  takes 
place  during  the  discharge  of  the  accumulator  is 

(1)  Pb  +  2Pb02  +  2H2S04  =  2PbS04  +  2H2O 
or,  written  in  ionic  form, 

(2)  Pb  +  Pb  •  •  •  •  =  2Pb  •  •. 

Lead  peroxide,  or  rather  its  hydroxide,  can  be  regarded  as 
the  hydroxide  of  tetravalent  lead.  In  presence  of  lead  per- 
oxide as  solid  phase,  and  therefore  at  the  positive  electrode  of 
the  accumulator,  the  concentration  of  the  tetravalent  lead 

ions,  [Pb ],  will  be  determined  by  the  solubility  of  lead 

peroxide. 

We  may  regard  reaction  (2)  as  made  up  of  two  reactions, 
one, 

Pb  ->  Pb  •  •, 
localised  at  the  negative  electrode,  and  the  other, 

Pb-----»Pb--f 
localised  at  the  positive  electrode.     The  potential  difference 


ELECTEOMOTIVE  FORCE—  EXAMPLES       173 

between  the  two  electrodes  must  be  equal  to  the  difference  of 
the  single  potential  differences  between  the  separate  elec- 
trodes and  the  solution,  therefore,  from  (5)  and  (10), 


(3)       =  c0  +         log,  [Pb  '  ']  -  «  o 

The  lead  peroxide  electrode  is  here  regarded  as  an  oxidation 
electrode.  E,  e0  and  e'0  are  given  ;  the  required  concentra- 
tion of  plumbic  ions,  [Pb  •'•'],  may,  therefore,  be  determined, 
if  [Pb  •  •]  can  be  calculated.  This  can  be  done  from  the  given 
solubility,  a,  of  lead  sulphate,  with  which  the  solution  is 
saturated.  In  a  saturated  pure  aqueous  solution  of  lead 
sulphate,  [Pb  •  •]  [SO4"]  =  L  =  a2,  if  we  assume  that  the 
dissociation  is  complete.* 

In  molecular  N-H2S04  the  concentration  of  the  SO4"- 
ions  would  be  [S04"]  =  1,  if  the  dissociation  of  the  acid  were 
complete  ;  if  the  degree  of  dissociation  is  assumed  to  be 

approximately  50  per  cent.,  [SO/']  =  0-5  and  [Pb  •  •]  =  _£L 
Therefore,  from  (3), 


and,  changing  to  common  logarithms, 


The  sign  of  E  in  the  above  equation  must  be  taken  as 
negative,  because,  according  to  (3),  E  is  the  potential  differ- 
ence between  the  lead  electrode  and  the  lead  peroxide  electrode, 
and  as  the  lead  electrode  is  the  negative  pole,  E  =  -  1-88. 

-  2  x  7-5  =  -  16-35 


.'.  [Pb  ----  ]  =  4*5  x  10  ~  17  gram-ion  per  litre. 
(2)  During  the  discharge  the  concentration  of  the  sulphuric 
acid  diminishes,  and,  according  to  reaction   (1),  to   such  an 

*  This  assumption  is,  however,  only  approximately  correct. 


174       ELECTEOMOTIVE  FOECE—  EXAMPLES 

extent  that  for  every  equivalent  of  lead  which  goes  into 
solution  at  the  negative  electrode  one  molecule  of  H2S04 
is  deposited  as  PbSO4.  By  a  discharge  of  one  ampere-hour, 


therefore,          c  =  0-037  molecules  of  H2S04  are  so  deposited 
yoo  Ttvj 

and  removed  from  the  solution.  The  100  c.c.  of  molecular 
N-H2SO4,  which  originally  contained  Ol  molecule  H2SO4, 
contain  after  the  discharge  only  Ol  -  0*037  =  0-063  mole- 
cules. The  concentration  of  the  H2S04  is,  therefore,  only 
0-63  molecular  normal  after  the  discharge.  If  we  assume 
that  the  degree  of  dissociation  of  this  acid  is,  approximately, 
60  per  cent.,  the  concentration  of  the  SO4"-ions  is 
[SO/']  =  0-63  x  0-6  =  0-38, 

and  the  concentration  of  the  Pb  •  --ions  [Pb  •  •]  =    a     . 

0*38 

The  concentration  of  the  Pb  ----  -ions  also  changes  during 
the  discharge  with  the  change  in  concentration  of  the  acid. 

In  the  solution,  which  is  saturated  with  lead  peroxide,  the 
equation 


must  always  be  satisfied,  where  L1  denotes  the  solubility- 
product  of  plumbic  hydroxide,  and,  since  [H'][OH'J  =  Klt  we 
obtain 


In  two  solutions,  which  we  may  distinguish  by  the  suffixes 
1  and  2,  we  have,  therefore, 

[Pb--  ••],_[!!•],« 

••••],    [H-y 


Before  the  discharge  the  H'-ion  concentration  of  the  H2S04 
is  [H-]t  =  2  x  0*5  =  1,  if  we  assume  the  dissociation  to  take 
place  only  according  to  the  equation  H2SO4  =  2H-  +  SO4"  ; 
after  the  discharge  it  is  [H-]2  =  2  x  O38  =  0'76.  Therefore 


=  4-5  x  10  ~  17  x  0-33  =  1-5  x  10  ~17  gram-ion  per  litre. 

(3)  The   E.M.F.  of   the  accumulator   after   the   discharge 
follows  from  the  equation  (cf.  (3)) 


ELECTEOMOTIVE  FOECE— PROBLEMS       175 


r2PA  '  2F1    •  [PFT 

^  =  €'°  "  €°  +  Wlo&  [pb  ""I-  ^lo&  [Pb  •  •] 

=  1-80  +  0-12  +  0-0295  log  1-5  x  10  - 17  -  0-059  log 1>58  x  1Q~8 

0*38 

=  1-92  -  0-496  +  0-436 
=  r86  volt. 

E.M.F.— Problems  for  Solution 

PROBLEM  335. — The  equivalent  conductivity  at  18°  of  a 
0-02  equivalent  N-ZnCl2  solution  is  94  r.o.,  and  the  equiva- 
lent conductivity  of  ZnCl2  at  infinite  dilution  is  112  r.o.  The 
normal  potential  of  Zn  (referred  to  the  N-H-  electrode  as 
zero)  is  -  0-770  volt.  What  is  the  potential  of  Zn  against 
a  0-02  equivalent  N-ZnCl2  solution  at  18°? 

Ans.    -  0-830  volt. 
PROBLEM  336.— At  25°  the  E.M.F.  of  the  cell 

Zn/0-5  mol.  N-ZnSO4/0-05  mol.  N-ZnSO4/Zn 

is  0-018  volt.  Neglecting  the  diffusion-potential  at  the  junc- 
tion of  the  electrolytes,  and  assuming  the  dilute  solution  to 
be  35  per  cent,  dissociated,  calculate  the  degree  of  dissociation 
of  the  concentrated  solution. 

Ans.    0-142. 

PROBLEM  337. — The  E.M.F.  of  the  cell 
Ag/0-1  N-AgNO3/saturated  NH4NO3/0'01  N-AgN03/Ag 

is  0-0556  volt  at  25°.  At  25°  the  specific  conductivity  of 
0-1  N-AgNO3  is  109-3  x  10  ~4  r.o.,  and  of  O'Ol  N-AgNO3 
12-53  x  10  ~4  r.o.  On  the  assumption  that  the  conductivity 
is  a  true  measure  of  the  ionic  concentration,  calculate  the 
E.M.F.  of  the  Ag  concentration -cell  given  above,  and  com- 
pare with  the  value  found. 

Ans.   Calculated  value  =  0-0555  volt. 

PROBLEM  338.— The  E.M.F.  of  a  Daniell  cell  in  which  the 
Cu  •  •-  and  Zn  •  --ion  concentrations  are  equal  is  1-1  volt  at 
18°.  What  is  the  E.M.F.  at  18°  of  a  Daniell  cell  in  which 


176       ELECTROMOTIVE  FORCE— PEOBLEMS 


the  Cu  •  --ion  concentration  is  0-0005  and  the  Zn 
centration  is  0-5  gram-ion  per  litre  ? 

Ans.    1-013  volt. 
PROBLEM  339.— At  25°  the  E.M.F.  of  the  cell 


•-ion  con- 


Pb     0.01  mol.  N-Pb(NO3)2 


saturated 
NH4N03 


is  -  0-469  volt,  and  that  of  the  cell 


Pb    001'  mol.   N-Pb(C103)2 


N-calomel 
electrode 


N-calomel 
electrode 


is  -  0-463  volt.  The  NH4NO3  solution  eliminates  the  dif- 
fusion-potentials. If  the  Pb(C103)2  is  assumed  to  be  com- 
pletely dissociated,  what  is  the  degree  of  dissociation  of  the 
Pb(N03)2? 

Ans.    62*1  per  cent. 
PROBLEM  340.— The  E.M.F.  of  the  cell 

Pb     0-01   mol.  N-Pb(N03)2     NH4N03 

is  -  0-469  volt  at  25°.  The  Pb(NO3)2  is  62  per  cent,  dissoci- 
ated. What  is  the  normal  potential  of  Pb  referred  to  the 
N-calomel  electrode  as  zero  ? 

Ans.    -  0-405  volt. 

PROBLEM  341.— At  25°  the  E.M.F.  of  the  cell 
saturated 


N-calomel 
electrode 


* 

Ag 


f\  i  XT  A   AT  r\ 
0-lN-AgN0 


0-1  N-calomel  electrode 


is  0-396  volt,  and  that  of  the  cell 


saturated  solution 
of  Ag(C2H302) 


saturated      O'l  N-calomel 
NH4NO3  electrode 

is  0-383  volt.  0-1  N-AgNO3  is  82  percent,  dissociated. 
Calculate  (a)  the  Ag  •-  ion  concentration  in  the  saturated 
solution  of  silver  acetate  and  (6)  the  degree  of  dissociation  of 
the  saturated  solution,  given  that  the  solubility  of  the  silver 
acetate  at  25°  is  0-0664  gram-molecule  per  litre. 

Ans.  (a)  0-04938  gram-ion  per  litre,  (b)  0'7435. 
PROBLEM  342.— The  E.M.F.  of  the  cell 
Ag/AgCNS  in  0-1  N-KCNS/AgBr  in  0-1  N-KBr/Ag 


ELECTROMOTIVE  FORCE— PROBLEMS   177 

is  0'015  volt  at  25°.  The  solubility  of  AgBr  in  pure  aqueous 
solution  is  7'2  x  10~7  gram- molecule  per  litre.  Neglecting 
the  diffusion-potential  and  assuming  that  all  the  salts  are 
completely  dissociated,  calculate  (a)  the  solubility- product 
and  (b)  the  solubility  of  AgCNS  at  25°. 

Ans.  (a)  0-93  x  10- 12,  (b)  0;96  x  10~6  gram-molecule 
per  litre. 

PROBLEM  343. — The  E.M.F.  of  the  cell 


N-AgNOi 


Ag 


is  -  0-198  volt  at  25°.  Neglecting  the  diffusion-potential, 
and  taking  the  01  N-AgNO3  as  83  per  cent,  dissociated 
and  the  O'Ol  N -K2Ca04  as  completely  dissociated,  calculate 
(a)  the  solubility-product  and  (b)  the  solubility  of  silver 
oxalate. 

Ans.  (a)  1-3  x  10~n,  (b)  1-48  x  10- 4  gram-molecule 
per  litre. 

PROBLEM  344.— What  is  the  E.M.F.  at  25°  of  the  cell 
H^O-S  N-HC1/01  N-NaOH/H2 

if  the  H2  at  each  electrode  is  under  atmospheric  pressure 
and  if  the  diffusion-potential  is  neglected?  The  degree  of 
dissociation  of  0'5  N-HC1  =  0-87,  of  01  N-NaOH  =»  0'9 
and  the  ionic-product  of  water  is  1*2  x  10~14. 

Ans.  0-738  volt. 

PROBLEM  345.— What  is  the  E.M.F.  at  25°  of  the  cell 
H2/0-5  N-fonnic  acid/1  N-acetic  acid/H2 

if  the  diffusion-potential  is  neglected?  The  dissociation- 
constant  of  formic  acid  is  127  x  10 -5  and  of  acetic  acid 
1-8  x  10- 5. 

Ans.  0-0451  volt. 
PROBLEM  346.— The  E.M.F.  of  the  cell 


A     I     AgCl  in 
Ag    01  N-KC1 


saturated 
NH4NO3 


01  N-AgNO,     Ag 


is  -  0-450  volt  at  25°.     01  N-KC1  is  85  per  cent,  dissoci- 
ated  and   0-1    N-AgNO3   82   per   cent.      Calculate   (a)   the 
solubility-product  and  (b)  the  solubility  of  AgCl. 
12 


178   ELECTROMOTIVE  FORCE— PROBLEMS 

^        Ans.    (a)  1-65  x  10  -  10,  (b)  1-28  x  10  -  5  gram-molecule 

per  litre. 

PROBLEM  347.— The  E.M.F.  of  the  cell 

Cu/CuS04  +  100  H2O/ZnS04  +  100  H2O/Zn 
is  1-0960  volt  at  0°  C.  and  1-0961  volt  at  3°  C.     What  is  the 
heat  of  the  reaction  taking  place  in  the  cell  ? 

Ans.   63390  cals. 
PROBLEM  348.— The  E.M.F.  of  the  combination 

Cl2/molten  PbCl2/Pb 
is  given  by  the  formula 

E  =  1-263  -  [0-000679  (t  -  498)]  volts, 

where  t  is  the  temperature  C.    Calculate  the  heat  of  formation 
of  lead  chloride  at  498°  C. 

Ans.   82500  cals. 

PROBLEM  349. — At  18°  the  potential  of  a  Cu  electrode 
against  a  0'005  mol.  N-Cu(NO3)2  solution  is  0'266  volt, 
referred  to  the  N-H-  electrode  as  zero.  Assuming  that  the 
Cu(NO3)2  solution  is  completely  dissociated,  calculate  the 
normal  potential  of  Cu  against  Cu  •  --ions. 

Ans.   0-333  volt. 

PROBLEM  350. — The  normal  potential  of  Zn  referred  to  the 
N-H-  electrode  as  zero  is  -  0*770  volt  and  of  Cu  0-329 
volt.  When  excess  of  Zn  is  added  to  a  solution  of  CuSO4 
the  Zn  displaces  the  Cu  till  equilibrium  is  reached.  What 
is  the  ratio  of  the  concentration  of  the  Zn  •  •-  to  the  Cu  •  *-ions 
at  equilibrium  ?  (At  equilibrium  the  potentials  of  the  Zn 
and  Cu  against  the  solution  are  equal.) 

Ans.    [Zn  •  -]/[Cu  •  •]  =  7'94  x  1037. 

PROBLEM  351. — The  normal  potential  of  Ag  referred  to  the 
N-calomel  electrode  as  zero  is  0-488  volt  at  18°.  Taking  the 
absolute  potential  of  the  calomel  electrode  as  0-56  volt  (Hg 
positive),  calculate  the  electrolytic  solution  pressure  of  Ag  in 
atmospheres. 

Ans.    2-04  x  10  ~17. 

PROBLEM  352. — When  metallic  copper  is  shaken  with  a 
solution  of  a  cupric  salt  in  absence  of  air  at  20°,  the  reaction 


BLECTEOMOTIVE  FOKCB— PKOBLEMS       179 

Cu  +  Cu  •  •  =  2  Cu  •  proceeds  till  equilibrium  is  established. 
The  concentrations  of  the  Cu  •  •-  and  Cu  --ions  are  then  such 
that  [Cu  •  -]/[Cu  -]2  =  2-02  x  104.  If  the  normal  potential  of 
Cu  against  Cu  •  --ions  is  0*329  volt  (hydrogen  standard),  what 
is  the  normal  potential  of  Cu  against  Cu  '-ions  ? 

Ans.   0-454  volt. 

PEOBLEM  353. — When  CuCl  is  dissolved  in  KC1  solutions 
a  complex  salt  of  the  formula  (KCl)n*(CuCl)1D  is  formed,  which 
dissociates  giving  the  ions  K  •  and  CumCl'm  +  u.  |Two  very 
dilute  solutions  of  CuCl  in  0*1  N-KC1  were  prepared,  the 
one  containing  4  times  as  much  CuCl  as  the  other.  The 
E.M.F.  of  the  cell 


Cu 


dilute  solution  of 
CuCl  in  KC1 


cone,  solution  of 
CuCl  in  KC1 


Cu 


was  0-0351  volt  at  18°.  On  the  assumption  that  practically 
the  whole  of  the  dissolved  CuCl  is  present  as  complex  salt, 
and  that,  on  account  of  the  very  large  excess  of  KC1  over 
CuCl,  the  Cl'-ion  concentration  in  the  two  solutions  is  the 
same,  and  that  the  dissociation  of  the  KC1  and  of  the  com 
plex  salt  is  complete,  calculate  the  value  of  m. 

Ans.   0-995  (.-.  1,  as  it  must  be  a  whole  number). 

PROBLEM  354  (cf.  preceding  problem). — Two  solutions  of 
KC1,  one  0-21  N  and  the  other  O'l  N,  and  each  containing 
0-0002  gram-molecule  CuCl  per  litre,  were  prepared.  The 
E.M.F.  of  the  cell 

Cu/dilute  KC1  solution/cone.  KC1  solution/Cu 
was  0*0374  volt  at  18°.     Taking  the  value  of  m  found  above 
and  on  the  assumptions  made  in  the  preceding  problem,  cal- 
culate the  value  of  n. 

Ans.  ^L-n  -  2,  .-.  n  -  1. 
m 

PROBLEM  355.-— What  is  the  total  E.M.F  at  18°  of  the  eel 

Hj/0-1  N-HCl/0-001  N-HC1/H2 

if  the  H2  at  each  electrode  is  at  atmospheric  pressure? 
The  0-1  N-HC1  is  92  per  cent,  dissociated  and  the  O'OOl 
N-HC1  is  completely  dissociated.  The  ionic  conductivity 
of  H-  is  318  and  of  Cl'  65-4. 

Ans.   0-039  volt, 


180   ELECTKOMOTIVE  FORCE— PEOBLEMS 

PROBLEM  356. — A  saturated  solution  of  AgN02  contains 
0-0265  gram-molecule  per  litre  at  25°.  The  E.M.F.  of  the 
cell 


Ag 


AgN03  solution  containing 


0-0224  gram-ion  Ag  •  per  litre  of  AgN02 


saturated  solution 


Ag 


is  0-011  volt  at  25°.  Calculate  (a)  the  solubility  -  product 
of  AgNO2,  (b)  the  degree  of  dissociation  of  the  saturated 
solution. 

Ans.  (a)  2-13  x  10  ~4,  (b)  0'551. 

PROBLEM  357. — When  AgNO2  is  dissolved  in  KNO2  solu- 
tions the  complex  anion  Agm(NO'2)u  is  formed.  The  E.M.F. 
at  25°  of  the  cell 


0-584  N-KNO 
0-05  N-AgNO 


0-584  N-KNO^ 
0  025  N-AgN02 


is  0-0170  volt,  and  of  the  cell 


Ag 


0-379  N-KN02 
0-025  N-AgNO 


0-219  N-KN02      I  A 
0-025  N-AgN02    I  Ag 


is  -  0-0295  volt.  Assuming  that  practically  the  whole  of 
the  dissolved  AgN02  exists  as  complex  ion,  and  that  the 
NO'2-ioQ  concentrations  are  proportional  to  the  KNO2  con- 
centrations, calculate  the  values  of  m  and  n  and  hence  the 
formula  of  the  complex  ion. 

Ans.  m  =  1-04(.'.  1),  ~  =  2-1,  .-.  n  =  2  and  formula  of  com- 
plex ion  is  Ag(NO2)2. 

PROBLEM  358. — What  is  the  diffusion-potential  at  18°  at 
the  junctions 

(a)  0-001  N-HCl/0-001  N-KC1  and 

(b)  0-001  N-KCl/0-001  N-KOH, 

if  all  the  electrolytes  are  assumed  to  be  completely  dissoci- 
ated ?  The  ionic  conductivities  of  H-,  K  *,  01'  and  OH'  are 
318,  64-9,  65-4  and  174  respectively  at  18°. 

Ans.    (a)  0-0272  volt  (HC1  negative),  (b)  0'0153  volt  (KC1 
negative). 

PROBLEM  359  (cf.  preceding  problem).— What  is  the  total 
E,M.F.  at  16°  of  the  cell 


ELECTROMOTIVE  FORCE— PROBLEMS       181 

H2/0-001  N-HCl/0-001  N-KCl/0-001  N-KOH/H2 

if  the  H2  at  each  electrode  is  under  atmospheric  pressure  and 
all  the  electrolytes  are  completely  dissociated  ?  The  ionic 
product  of  water  is  1'2  x  10  ~ 14. 

Ans.   04199  volt. 
PROBLEM  360.— The  E.M.F.  of  the  cell 

Cu/N-CuS04/N-ZnS04/Zn 

is  ri  volt  at  18°.  What  is  the  maximum  work  (a)  in  joules, 
(b)  in  calories  obtainable  at  18°  by  the  reversible  displacement 
of  Cu  by  Zn  according  to  the  equation 

CuS04  +  Zn  =  ZnS04  +  Cu? 

Ans.    (a)  212300,  (b)  50670. 
PROBLEM  361.— The  E.M.F.  at  25°  of  the  cell 


Pt 


0-0337  mol.  N-T1(N03)3 
0-0216  mol.  N-T1N03 
0-42  N-HNO, 


N-H- 


H2  at 
atmos.  pr. 


is  1-2008  volt.  Taking  the  potential  of  H2/N-H-  as  zero, 
calculate  the  normal  potential  of  the  thallic-thallous  ion  elec- 
trode. Assume  that  the  concentrations  of  Tl  •  •  •-  and  Tl  --ions 
are  proportional  to  the  concentrations  of  the  corresponding 
nitrates  and  neglect  the  diffusion-potential.  The  HNO3  is 
added  to  prevent  hydrolysis  of  the  T1(NO3)3. 

Ans.   1-195  volt. 

PROBLEM  362  (cf.  preceding  problem). — What  is  the  ratio 
of  [Tl  •  •  •]  to  [Tl  •]  at  which  the  potential  of  a  thallic-thallous 
ion  electrode  is  zero  (referred  to  the  normal  hydrogen  elec- 
trode as  standard)  ? 

Ans.    [Tl  •  •  -]/[Tl  •]  =  6-1  x  10  ~42. 


CHAPTER  XI 
DIFFUSION— RADIOACTIVITY 

Diffusion 

WHEN  a  dissolved   substance  diffuses  from  a   place  •  of 
higher  concentration  c  to  a  place  of  lower   concen- 
tration c  -  dc,  the  amount  which  diffuses  in  unit  time  through 
unit  cross-section  is  proportional  to  the  concentration-gradient 

--,  and  is,  therefore,  equal  to  D-^>  (Fick's  law),  where  dx  is 

the  thickness  of  the  diffusion-layer.  The  factor  of  propor- 
tionality D  is  called  the  diffusion-coefficient  of  the  dissolved 
substance. 

Diffusion— Examples 

PROBLEM  363. — An  electrolytic  trough  contains  a  small 
rotating  platinum  cathode  and  a  large  platinum  anode.  The 
electrolyte  is  sulphuric  acid  which  contains. r  0-0005  gram  of 
iodine  per  c.c.  At  all  potentials  between  0'2  and  1-0  volt  a 
constant  current  (residual  current)  of  C  =  O'OOl  ampere 
passes  through  the  cell.  What  is  the  thickness  8  of  the  layer 
which  clings  by  adhesion  to  the  rotating  cathode,  if  the  area 
of  the  cathode  surface  is  A  =  0'2  sq.  cm.  and  the  diffusion- 
coefficient  of  iodine  is  D  =  0*6  cm.2  per  day  ?  What  is  the 
value  of  the  residual  current  if  the  sulphuric  acid  contains 
0-0008  gram  of  iodine  per  c.c.  ? 

SOLUTION  363. — Every  electrolyte  is  decomposed  by  the 
passage  of  the  electric  current.  For  every  electrolyte,  how- 
ever, a  definite  minimum  E.M.F.  is  necessary  for  the  free 
separation  of  the  decomposition  products  at  the  electrodes. 
This  minimum  E.M.F.  is  called  the  decomposition-potential 
of  the  electrolyte.  If  the  E.M.F.  applied  is  less  than  the 
decomposition-potential,  a  current  can  pass  only  if  the  de- 

182 


DIFFUSION— EXAMPLES  183 

composition  -  products  at  the  electrodes  are  continuously 
removed  by  inter-action  with  some  substance  present  in  the 
solution.  In  the  present  case,  for  example,  the  hydrogen 
evolved  at  the  cathode  is  "  depolarised  "  by  the  iodine  in  the 
solution  with  formation  of  hydrogen  iodide.  The  velocity 
with  which  this  reaction  takes  place  determines  the  strength 
of  the  current  which  flows  through  the  electrolyte  when  an 
E.M.F.  less  than  the  decomposition  -  potential,  is  applied. 
According  to  Nernst's  theory,  the  velocity  of  the  chemical 
reaction  between  iodine  and  hydrogen  at  the  electrode  is  very 
great,  whilst  the  diffusion  of  the  iodine  in  the  solution,  by 
which  the  supply  of  iodine  at  the  electrode  is  maintained, 
takes  place  slowly.  The  effective  velocity  of  the  depolarising 
reaction  is,  therefore,  determined  by  the  velocity  of  diffusion 
of  the  iodine  to  the  electrode. 

When  the  electrode  of  area  A  is  rotated,  a  layer  of  solution 
is  formed,  which  clings  to  the  surface  of  the  electrode  in  con- 
sequence of  adhesion  and  rotates  with  it.  The  concentration 
of  the  iodine  in  this  layer  diminishes  regularly  from  the  value 
c  (grams  per  c.c.),  the  concentration  of  the  iodine  in  the  rest 
of  the  solution,  to  the  value  0  at  the  electrode.  The  thick- 
ness of  this  adhesion  layer  depends  on  the  velocity  of  rotation. 
If  8  is  the  thickness  of  the  layer  for  a  definite  velocity  of 
rotation,  then,  according  to  Fick's  law,  there  diffuses  to  the 

electrode  in  the  time  dt  the  quantity  of  iodine  DA\--~~     '  dt 


grams,  where  D  denotes  the  diffusion-coefficient  of  iodine. 
If  M  is  the  64iiivalent  weight  of  iodine,  the  number  of 
equivalents  of  hydrogen  depolarised  by  the  diffused  iodine  is 

DAc 

— —  dt.     In  the  time  dt  the  current  C  liberates  the  quantity 

of  hydrogen  C '  c  •  dt,  where  c  denotes  the  quantity  liberated 
by  unit  current  in  unit  time,  that  is,  by  unit  quantity  of 
electricity.  In  the  stationary  condition,  therefore, 

~      .  ,f      DAc, 

"~8Mdt> 
or  the  required  thickness 

=  3f6«  " 

If  C  is  given  in  amperes  and  if  e  denotes  the  number  of 
equivalents  of  hydrogen  liberated  by  one  ampere  in  one 


184  DIFFUSION— EXAMPLES 

second,  that  is,  by  one  coulomb  of  electricity,  then  D  must  be 
expressed  in  cm.2/second. 
In  the  present  example 

)t6     =7-0  x  10-6cm.2/sec.,^  =0-2sq.cm.,c  =  0-0005 


gm.  per  c.c.,  C  =  O'OOl  amp.,  M  =  127,  €  =  ---,  equiv., 


.x       7-0  x  10-"  x  0-2  x  0-0005  x  96540 

127  x  0-001 
=  5*3  x  10  ~4  cm. 

If  the  concentration  of  iodine  in  the  solution  is  c'  instead 
of  c,  then,  if  C'  is  the  residual  current  in  this  case, 


r,,        7,       DAc'  7jt 

0     -.'««-     -gy-dt, 


and 


C'=  DAc> 


=  7-0  x  10 "6  x  0-2  x  0-0008  x  96540 

5-3  x  10-*  x  127 
=  0*0016  amp. 

PEOBLEM  364  (cf.  preceding  problem). — If  the  residual 
current  in  the  preceding  problem  is  C  =  0'05  ampere  when, 
instead  of  iodine,  the  sulphuric  acid  contains  0-01  gram  of 
bromine  per  c.c.,  what  is  the  diffusion-coefficient  of  bromine  ? 

SOLUTION  364.— If,  instead  of  iodine,  the  solution  con- 
tains bromine  of  concentration  c  =  O'Ol,  and  if  the  re- 
sidual current  is  C  =  0'05  ampere,  the  equation 

C,=    DAc 


applies  in  this  case  too,  if  D  is  the  diffusion-coefficient   and 
M  the  equivalent  weight  of  bromine.     We  obtain,  therefore, 

0-05  x  5-3  x  10  ~4  x  80 


Ac  0-2  x  0-01  x  96540 

=  1-1  x  10  ~5  cm.2/second, 
=  0*95  cm.2/day. 


EADIOACTIVITY— EXAMPLES  185 

Radioactivity 

A  radioactive  substance  gradually  loses  its  activity  C  accord- 
ing to  the  law  (Rutherford) 

C-Cf-u, 

or 

log,  G  =  log,  G0  -  Xt, 

G0  is  the  initial  activity,  C  the  activity  after  time  £,  e  the  base 
of  the  natural  logarithms  and  A.  a  constant  for  the  particular 
substance. 

Radioactivity — Example 

PROBLEM  365. — In  a  space  filled  with  radium  emanation 
the  current  strength  under  the  influence  of  a  strong  potential 
difference,  which  is  sufficient  to  produce  a  saturation  current, 
is,  at  a  definite  point  of  time,  G0  =  35-4  (in  arbitrary  units). 
After  tl  =  12  hours  the  current  strength  has  diminished  to 
G!  =  32-4,  and  after  t,  =  200  hours  to  Ga  =  8'10.  After 
what  time  has  the  current  strength  the  value  G3  =  10,  and 
after  what  time  has  it  sunk  to  half  its  original  value  ? 

SOLUTION  365. — The  current  strength  at  any  time  is  pro- 
portional to  the  amount  of  radium  emanation  present  at  that 
time.  The  decomposition  of  the  radium  emanation  follows 
the  exponential  law 


Taking  logarithms  of  both  sides  we  obtain 

The  value  of  \  may  be  obtained  from  each  of  the  equations 

and 

log  C,  =  log  C0  -  ***. 

A  o 

From  the  former  we  obtain 

2-3  log  S)     2-3  log  ^ 


A  =  Ll  =  °**  =  2-3  x_uuouu  =  0.00738 

^  12  12 


12 


186  RADIOACTIVITY—  EXAMPLES 

and  from  the  latter 


The  mean  value  of  X  is.  therefore,  0-00738. 

The  time,  t3,  after  which  the  current  strength  has  the  value 
C2  =  10,  is  obtained  from  the  equation 

2-3  log    >     2-3  log 


and  the  time,  t4,  after  which  the  current  strength  has  sunk 
to  half  its  original  value,  from  the  equation 


=  2-3  log  2      2-3  x  0-301 

--  =  -0-00738-  =  93'8 


INDEX 

Theoretical  matter  ig  indicated  by  ordinary  type,  solved  problems  by 
italics,  and  problems  for  solution  by  thick  type. 

Affinity  of  a  reaction,  67,  153 ;  81, 156,  162  ;  96,  181. 
Birnolecular  reaction,  velo3ity  of,  54  ;  56 ;  61. 
Boiling-point,  elevation  of,  29  ;  31 ;  39. 
Degree  of  association  of  liquids,  41 ;  41 ;  43. 

dissociation  of  electrolytes,  103  ;  27 1  29,  32,  105-117,  155 ;  12 

32-40,  136-142,  175-181. 
Density,  14  ;  14  ;  17. 

-  of  mixtures,  14  ;  14 ;  17. 
Diffusion,  182 ;  182. 
Diffusion-potential,  151 ;  154  ;  179-181. 
Dilution  law,  104;  105-117;  136-142. 

Dissociation-constant  of  electrolytes,  104;  105-117 ;  136-142. 
Dissociation  of  gases,  7  ;   7 ;  11. 
Electrode  potential,  149  ;  154-175;  175-181. 
Electrolytic  dissociation,  103-105;  105-117  ;  136-142. 
Electromotive  force,  149  ;  154-175;  175-181. 

of  concentration-cell,  150;  154  ;  175. 

—  galvaaic  element,  151 ;  154  ;  175. 
Equilibrium-constant,  62  ;  70-88 ;  91-96. 

and  temperature,  65  ;  70-88 ;  91-96. 

Equivalent  conductivity.  103  ;  105  ;  136-142. 

Faraday's  laws,  3*00  ;  102  ;  134. 

Fick's  law  of  diffusion,  182 ;  182. 

Free  energy,  change  of,  67,  153  ;  81, 166,  162;  96,  181. 

Freezing-point,  lowering  of,  28  ;  29 ;  35. 

Gas  laws,  1 ;  3 ;  9. 

Gibbs-Helmholtz  equation,  153 ;  162  ;  178. 

Heating  effect  of  current,  100  ;  101. 

Heat  of  reaction,  46;  47 ;  49. 

Hess's  law,  45 ;  47 ;  49. 

Hydrolysis  of  salts,  128  ;  130 ;  144-148. 

Ionic  conductivity,  104;  105-117;  136-142. 

Mass-action,  law  of,  b'2 ;  70-88  ;  91-96. 

Maximum  work,  67,  153;  81,  156, 162;  96,  181. 

Mixture  formula,  14,  20 

Migration  numbers,  117  ;  119 ;  135. 

Molecular  conductivity,  104 ;  136-142. 

—  elevation  of  boiling-point,  29 ;  31  ;  39. 

187 


188  INDEX 

Molecular  lowering  of  freezing-point,  28  ;  29 ;  35. 
"  —  weight  from  boiling-point,  29  ;  31 ;  39. 
—  freezing-point,  28  ;  29;  35. 

' vapour-pressure,  24  ;  25  ;  32. 

* of  liquids,  41;  41;  43. 

Monomolecular  reaction,  velocity  of,  53  ;  55  ;  60 
Normal  potential,  150  ;  154-175  ;  175-181 
Ohm's  law,  100 ;  101. 
Osmotic  pressure,  2  ;  9 ;  12. 
Oxidation-reduction  potential,  152  ;  164  ;  181 
Partition  law,  67  ;  89 ;  96. 
Quantity  of  electricity,  100. 
Radioactivity,  185 ;  185. 
Refractivity,  19  ;  20  :  21. 

—  of  mixtures,  20  ;  21 ;  22. 
Solubility,  121 ;  122 ;  142-146. 

—  of  gases,  68 ;  90 ;  98. 

Solubility-product,  121;  122,154-175;  142-146,  176-180 
Specific  conductivity,  103  ;  705,  122 ;  136-144 

--  volume,  14;  14;  17. 

of  mixtures,  14  ;  14 ;  17. 

Surface  tension,  41 ;  41  ;  43. 
Thermochemistry,  45  ;  47  ;  49. 
Transport  numbers,  117  ;  119 ;  135. 
Vapour-pressure,  lowering  of,  24;  25;  32. 

—  and  temperature,  25  ;  25  ;  32. 
Velocity  of  migration,  10 1 ;  139. 

reaction,  53;  55, 108;  60,  147. 

and  temperature,  58.  a  £ 


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